Information

How can both strands of DNA code for proteins with similar functions?


It's not clear from the question but for example:

AAAAAAA

TTTTTTT

The top strand would create a different protein than the bottom, and with the huge amount of nucleotide in a gene, I think it's very unlikely that the same region on both gene could create proteins that are similar to each other, even though they're alleles of the same gene. This would also affect codominance/incomplete dominance, since I'm assuming that only one of the strand would create a functional protein.


You are confusing the the forward and reverse strands of DNA as two alleles. They are not. Remember you have a double stranded DNA from each parent. Each of those two pieces of double stranded DNA represent alleles for a given locus.

For the example you gave, you can have different genes that overlap the same region of DNA. One is on one strand and a different gene is on the other. The DNA is read 5' to 3' so we think of one strand as forward and one strand as reverse.

See embedded pic for examples from this publication about different kinds of overlap and whether they are conserved across species.


I think it's very unlikely that the same region on both gene could create proteins that are similar to each other, even though they're alleles of the same gene

I think your reasoning is faulty here. It's not that the two alleles are in the same chromosome and each in one strand. Each allele is in a different chromosome (one from each parent).

Overlapping or nested genes are a different topic altogether.


In general, protein-coding genes don't overlap, so the problem you've identified doesn't come up.

And no, complementary strands are not different alleles. If that were so, bacteria and gametes would have two alleles of every gene, because they each contain one double stranded copy of the genome.

You have two copies of each double-stranded chromosome, that is why you have two alleles for each gene.


Usually not the whole gene overlaps, and it's not usually in-frame, this makes it a lot easier.

You also have some flexibility because of the redundant codons (20 AA for 64 codons), so you can more or less change every 3rd base to fit the "other" gene.

"Similar" could mean anything, possibly referring to two proteins that are not even 20% identical. At 20% identity you can still be sure that that enzyme is of the same class, catalyzing the same reaction. It might accept different substrates, but it could also still accept the same substrates.

Together, this means that on the non-coding strand of every gene you can put a gene coding for a protein that's more or less doing the same thing.


How can both strands of DNA code for proteins with similar functions? - Biology

21. DNA and Biotechnology

In the previous chapter, we learned about the chromosomal basis of inheritance. In this chapter, we become familiar with the structure and function of DNA and discover how this molecule is able to serve as the basis of our genetic inheritance as well as the source for the diversity of life on Earth. We learn that the importance of DNA on a personal level is that it directs the synthesis of specific polypeptides (proteins) that play structural or functional roles in our bodies. We then consider the technology that our understanding of DNA has already made available and what possibilities such technology may hold for the future.

Form of DNA

DNA is sometimes referred to as the thread of life—and a very slender thread it is. When DNA is unwound, it measures a mere 50-trillionths of an inch in diameter. If all the DNA strands in a single cell were fastened together end to end, the thread would stretch more than 5 ft in length. DNA might also be considered the thread that ties all life together, because the DNA of organisms ranging from bacteria to humans is built from the same kinds of subunits. The order of these subunits encodes the information needed to make the proteins that build and maintain life.

Deoxyribonucleic acid, or DNA, is a double-stranded molecule resembling a ladder that is gently twisted to form a spiral called a double helix, as shown in Figure 21.1. Each side of the ladder, including half of each rung, is made from a string of repeating subunits called nucleotides. You may recall from Chapter 2 that a nucleotide is composed of three subunits, including one sugar (deoxyribose, in DNA), one phosphate, and one nitrogenous base. DNA contains four types of nitrogenous bases: adenine (A), guanine (G), thymine (T), and cytosine (C). The sides of the ladder are composed of alternating sugars and phosphates the rungs consist of paired nitrogenous bases. The bases attach to each other according to the rules of complementary base pairing: adenine pairs only with thymine (creating an A-T pair), and cytosine pairs only with guanine (creating a C-G pair). Each base pair is held together by weak hydrogen bonds. The pairing of complementary bases is specific due to the shapes of the bases and the number of hydrogen bonds that can form between them. You may also recall from Chapter 2 that a molecule formed by the joining of nucleotides is called a nucleic acid. Thus, DNA is a nucleic acid.

· A DNA fingerprint is based on the sequence of bases in a person's DNA. Thus, a DNA fingerprint is unique to each person (except for those who happen to have identical siblings).

FIGURE 21.1. DNA is a double-stranded molecule that twists to form a spiral structure called a double helix.

Because base pairing is so specific, the bases on one strand of DNA are always complementary to the bases on the other strand. Thus the order of bases on one strand determines the sequence of bases on the other strand. For instance, if the sequence of bases on one strand were CATATGAG, what would the complementary sequence be? Remember, cytosine (C) always pairs with guanine (G), and adenine (A) always pairs with thymine (T). As a result, the complementary sequence on the opposite strand would be GTATACTC.

The DNA within each human cell has an astounding 3 billion base pairs. Although the pairing of adenine with thymine and cytosine with guanine is specific and does not vary, the sequence of bases throughout the length of different DNA molecules can vary in myriad ways. As we will see, genetic information is encoded in the exact sequence of bases.

Replication of DNA

For DNA to be the basis of inheritance, its genetic instructions must be passed from one generation to the next. Moreover, for DNA to direct the activities of each cell, its instructions must be present in every cell. These requirements dictate that DNA be copied before both mitotic and meiotic cell division (see Chapter 19). It is important that the copies be exact. The key to the precision of the copying process is that the bases are complementary.

The copying process, or DNA replication, begins when an enzyme breaks the weak hydrogen bonds that hold together the paired bases that make up nucleotide strands of the double helix, thereby "unzipping" and unwinding the strands. As a result, the nitrogenous bases on the separated regions of each strand are temporarily exposed. Free nucleotide bases, which are always present within the nucleus, can then attach to complementary bases on the open DNA strands. Enzymes called DNA polymerases link the sugars and phosphates of the newly attached nucleotides to form a new strand. As each of the new double-stranded DNA molecules forms, it twists into a double helix.

Each strand of the original DNA molecule serves as a template for the formation of a new strand. This process is called semiconservative replication, because in each of the new double-stranded DNA molecules, one original (parent) strand is saved (conserved) and the other (daughter) strand is new. Look at Figure 21.2. Notice the original (parental) strand of nucleotides in each new molecule of DNA. Complementary base pairing creates two new DNA molecules that are identical to the parent molecule.

FIGURE 21.2. DNA replication is called semiconservative because each daughter molecule consists of one “parental" strand and one “new" strand.

Gene Expression

The process of replication ensures that genetic information is passed accurately from a parent cell to daughter cells and from generation to generation. The next obvious question is, "How does DNA issue commands that direct cellular activities?" The answer is that DNA directs the synthesis of another nucleic acid— ribonucleic acid, or RNA. RNA, in turn, directs the synthesis of a polypeptide (a part of a protein) or a protein. The protein may be a structural part of the cell or play a functional role, such as an enzyme that speeds up certain chemical reactions within the cell.

Recall from Chapter 20 that a gene is a segment of DNA that contains the instructions for producing a specific protein (or in some cases, a specific polypeptide).1 The sequence of bases in DNA determines the sequence of bases in RNA, which in turn determines the sequence of amino acids of a protein. We say that the gene is expressed when the protein it codes for is produced. The resulting protein is the molecular basis of the inherited trait it determines the phenotype.

To more fully appreciate how gene expression works, we will consider each step in slightly greater detail.

Just as the CEO of a major company issues commands from headquarters instead of from the factory floor, DNA issues instructions from the cell nucleus and not from the cytoplasm where the cell's work is done. RNA is the intermediary that carries the information encoded in DNA from the nucleus to the cytoplasm and directs the synthesis of the specified protein.

Like DNA, RNA is composed of nucleotides linked together, but there are some important differences between DNA and RNA, as shown in Table 21.1. First, the nucleotides of RNA contain the sugar ribose, instead of the deoxyribose found in DNA. Second, in RNA the nucleotide uracil (U) pairs with adenine, whereas in DNA thymine (T) pairs with adenine (A). Third, most RNA is single stranded. Recall that DNA is a double-stranded molecule.

TABLE 21.1. Comparisons of DNA and RNA

Are composed of linked nucleotides

Have a sugar-phosphate backbone

Is a double-stranded molecule

Is a single-stranded molecule

Contains the bases adenine, guanine, cytosine, and thymine

Contains the bases adenine, guanine, cytosine, and uracil (instead of thymine)

Functions primarily in the nucleus

Functions primarily in the cytoplasm

The first step in converting the DNA message to a protein is to copy the message as RNA, by a process called transcription.

Three types of RNA are produced in cells. Each plays a different role in protein synthesis (Table 21.2). Messenger RNA (mRNA) carries DNA's instructions for synthesizing a particular protein from the nucleus to the cytoplasm. The order of bases in mRNA specifies the sequence of amino acids in the resulting protein, as we will see. Each transfer RNA (tRNA) molecule is specialized to bring a specific amino acid to where it can be added to a polypeptide that is under construction. Ribosomal RNA (rRNA) combines with proteins to form ribosomes, which are the structures on which protein synthesis occurs.

TABLE 21.2. Review of the Functions of RNA

Carries DNA's information in the sequence of its bases (codons) from the nucleus to the cytoplasm

Binds to a specific amino acid and transports it to be added, as appropriate, to a growing polypeptide chain

Combines with protein to form ribosomes (structures on which polypeptides are synthesized)

Transcription begins with the unwinding and unzipping of the specific region of DNA to be copied these actions are performed by an enzyme. The DNA message is determined by the order of bases in the unzipped region of the DNA molecule. One of the unwound strands of the DNA molecule serves as the template during transcription. RNA nucleotides present in the nucleus pair with their complementary bases on the template—cytosine with guanine and uracil with adenine (Figure 21.3). The signal to start transcription is given by a specific sequence of bases on DNA, called the promoter. An enzyme called RNA polymerase binds with the promoter on DNA and then moves along the DNA strand, opening up the DNA helix in front of it and then aligning the appropriate RNA nucleotides and linking them together the region of DNA that has been transcribed zips again after RNA polymerase passes by. Another sequence of bases on the DNA signals RNA polymerase to stop transcription. After transcription ceases, the newly formed strand of RNA, called the RNA transcript, is released from the DNA.

FIGURE 21.3. Transcription is the process of producing RNA from a DNA template.

Messenger RNA usually undergoes certain modifications before it leaves the nucleus (Figure 21.4). Most stretches of DNA between a promoter and the stop signal include regions that do not contain codes that will be translated into protein. These unexpressed regions of DNA are called introns, short for intervening sequences. The regions of mRNA corresponding to the introns are snipped out of the newly formed mRNA strand by enzymes before the strand leaves the nucleus. The remaining segments of DNA or mRNA, called exons for expressed sequences, splice together to form the sequence that directs the synthesis of a protein.

FIGURE 21.4. Newly formed messenger RNA is modified before it leaves the nucleus. Noncoding regions of DNA called introns are snipped out of the corresponding regions of mRNA molecule. Segments of mRNA that code for protein are then spliced together.

The newly formed mRNA carries the genetic message (transcribed from DNA) from the nucleus to the cytoplasm, where it is translated into protein at the ribosomes. Just as we might translate a message written in Spanish into English, translation converts the nucleotide language of mRNA into the amino acid language of a protein.

Before examining the process of translation, we should become more familiar with the language of mRNA.

The genetic code . To use any language, you must know what the words are and what they mean, as well as where sentences begin and end. The genetic code is the "language" of genes that translates the sequence of bases in DNA into the specific sequence of amino acids in a protein. We have seen that the sequence of bases in DNA determines the sequence of bases in mRNA through complementary base pairing. The "words" in the genetic code, called codons, are sequences of three bases on mRNA that specify 1 of the 20 amino acids or the beginning or end of the protein chain. All the codons of the genetic code are shown in Figure 21.5. For instance, the codon UUC on mRNA specifies the amino acid phenylalanine. (The complementary sequence on DNA would be AAG.)

FIGURE 21.5. The genetic code. Each sequence of three bases on the mRNA molecules, called a codon, specifies a specific amino acid, a start signal, or a stop signal.

If the sequence of bases following a start signal were AACUCAGCC, what amino acids would be specified?

Asparagine, serine, alanine

Look at the strand of mRNA in Figure 21.3. Notice that the codon at the end of the mRNA strand is ACG. Which amino acid does this specify? (Use Figure 21.5.)

The four bases in RNA (A, U, C, and G) could form 64 combinations of three-base sequences. The number of possible codons, therefore, exceeds the number of amino acids. As Figure 21.5 indicates, there are several sets of codons that code for the same amino acid. Note, too, that the codon AUG can either serve as a start signal to initiate translation or can specify the addition of the amino acid methionine to the growing protein chain, depending on where it occurs in the mRNA molecule. In addition, three codons (UAA, UAG, and UGA) are stop codons that signal the end of a protein and that do not code for an amino acid. If we think of the codons as genetic words, then a stop codon functions as the period at the end of the sentence.

Transfer RNA A language interpreter translates a message from one language to another. Transfer RNA (tRNA) serves as an interpreter that converts the genetic message carried by mRNA into the language of protein, which is a particular sequence of amino acids. To accomplish this conversion, a tRNA molecule must be able to recognize both the codon on mRNA and the amino acid that the codon specifies—in other words, it must speak both languages.

There are many kinds of tRNA—at least one for each of the 20 amino acids. Each type of tRNA molecule binds to a particular amino acid. Enzymes ensure tRNA binds with the correct amino acid. The tRNA then ferries the amino acid to the correct location along a strand of mRNA (Figure 21.6).

FIGURE 21.6. A tRNA molecule Is a short strand of RNA that twists and folds on Itself. The job of tRNA Is to ferry a specific amino acid to the ribosome and insert it in the appropriate position in the growing peptide chain.

How does the tRNA know the correct location along mRNA? The location is determined by a sequence of three nucleotides on the tRNA called the anticodon. In a sense, the anticodon "reads" the language of mRNA by binding to a codon on the mRNA molecule according to the complementary base-pairing rules. When the tRNA's anticodon binds to the mRNA's codon, the specific amino acid attached to the tRNA is brought to the growing polypeptide chain. For example, a tRNA molecule with the anticodon AAG binds to the amino acid phenylalanine and ferries it to the mRNA molecule, where the codon UUC is presented for translation. Phenylalanine will then be added to the growing amino acid chain.

Ribosomes . Ribosomes function as the workbenches on which proteins are built from amino acids. A ribosome consists of two subunits (small and large), each composed of ribosomal RNA (rRNA) and protein. The subunits form in the nucleus and are shipped to the cytoplasm. They remain separate except during protein synthesis. The role of the ribosome in protein synthesis is to bring the tRNA bearing an amino acid close enough to the mRNA to interact. As you can see in Figure 21.7, when the two subunits fit together to form a functional ribosome, a groove for mRNA is formed. Two binding sites position tRNA molecules so that an enzyme in the ribosome can cause bonds to form between their amino acids.

FIGURE 21.7. A ribosome consists of two subunits of different sizes. When the two subunits join together to form a functional ribosome, a groove for mRNA is formed. The ribosome has two binding sites for tRNA molecules. It also contains an enzyme that promotes the formation of a peptide bond between the amino acids that are attached to the tRNAs in the binding sites.

Protein synthesis . Translation—essentially, protein synthesis— can be divided into three stages: initiation, elongation, and termination.

1. During initiation, the major players in protein synthesis (mRNA, tRNA, and ribosomes) come together (Figure 21.8).

• Step 1: The small ribosomal subunit attaches to the mRNA strand at the start codon, AUG.

• Step 2: The tRNA with the complementary anticodon pairs with the start codon. The larger ribosomal subunit then joins the smaller one to form a functional, intact ribosome with mRNA positioned in a groove between the two subunits.

FIGURE 21.8. Initiation of translation

2. Elongation of the protein occurs as additional amino acids are added to the chain (Figure 21.9).

• Step 1: Codon recognition. With the start codon positioned in one binding site, the next codon is aligned in the other binding site.

• Step 2: Peptide bond formation. The tRNA bearing an anticodon that will pair with the exposed codon slips into place at the binding site, and the amino acid it carries forms a peptide bond with the previous amino acid with the assistance of enzymes.

• Step 3: Ribosome movement. The tRNA in the first binding site leaves the ribosome. The ribosome moves along the mRNA molecule, carrying the growing peptide chain and the remaining tRNA with its amino acid to the first binding site. This movement positions the next codon in the open site. An appropriate tRNA slips into the open site, and its amino acid binds to the previous one. This process is repeated many times, adding one amino acid at a time to the growing polypeptide chain.

FIGURE 21.9. Elongation of the polypeptide during translation

Many ribosomes may glide along a given mRNA strand at the same time, each producing its own copy of the protein directed by that mRNA (Figure 21.10). As soon as one ribosome moves past the start codon, another ribosome can attach. A cluster of ribosomes simultaneously translating the same mRNA strand is called a polysome.

FIGURE 21.10. A polysome is a group of ribosomes reading the same mRNA molecule.

3. Termination occurs when a stop codon moves into the ribosome (Figure 21.11).

• Step 1: Stop codon moves into ribosome. There are no tRNA anticodons that pair with the stop codons, so when a stop codon moves into the ribosome, protein synthesis is terminated.

• Step 2: Parts disassemble. The newly synthesized polypeptide, the mRNA strand, and the ribosomal subunits then separate from one another.

FIGURE 21.11. Termination of translation

Streptomycin is an antibiotic, a drug taken to slow the growth of invading bacteria and allow body defense mechanisms more time to destroy them. Streptomycin works by binding to the bacterial ribosomes and preventing an accurate reading of mRNA. Why would this process slow bacterial growth?

DNA is remarkably stable, and the processes of replication, transcription, and translation generally occur with amazing precision. However, sometimes DNA is altered, and the alterations can change its message. Changes in DNA are called mutations. One type of mutation occurs when whole sections of chromosomes are duplicated or deleted, as discussed in Chapter 20. Now that we are familiar with the chemical structure of DNA and how it directs the synthesis of proteins, we can consider another type of mutation—a gene mutation. A gene mutation results from changes in the order of nucleotides in DNA. Although a gene mutation can occur in any cell, the only way it can be passed on to offspring is if it is present in a cell that will become an egg or a sperm. A mutation that occurs in a body cell can affect the functioning of that cell and the subsequent cells produced by that cell, sometimes with disastrous effects, but it cannot be transmitted to a person's offspring.

One type of gene mutation is the replacement of one nucleotide pair by a different nucleotide pair in the DNA double helix. During DNA replication, bases may accidentally pair incorrectly. For example, adenine might mistakenly pair with cytosine instead of thymine. Repair enzymes normally replace the incorrect base with the correct one. However, sometimes the enzymes recognize that the bases are incorrectly paired but mistakenly replace the original base (the one on the old strand) rather than the new, incorrect one. The result is a complementary base pair consisting of the wrong nucleotides (Figure 21.12).

FIGURE 21.12. A base-pair substitution is a DNA mutation resulting when a base is paired incorrectly. This may change the amino acid specified by the mRNA and alter the structure of the protein.

Other types of gene mutations are caused by the insertion or deletion of one or more nucleotides. Generally, a mutation of this kind has more serious effects than does a mutation caused by substitution of one base pair for another. Recall that the mRNA is translated in units of three nucleotides (a unit called a codon). If one or two nucleotides are inserted or deleted, all the triplet codons that follow the insertion or deletion are likely to change. Consequently, mutations due to the insertion or deletion of one or two nucleotides can greatly change the resulting protein. A sentence consisting of three- letter words (representing codons) illustrates what can happen. Deleting a single letter from the sentence "The big fat dog ran" renders the sentence nonsensical:

Original: THE BIG FAT DOG RAN

After deletion of the E in THE: THB IGF ATD OGR AN

Regulating Gene Activity

At the time of your conception, you received one set of chromosomes from your father and one set from your mother. The resulting zygote then began a remarkable series of cell divisions—some of which continue in many of your body cells to this day. With each division, the genetic information was faithfully replicated, and exact copies were parceled into the daughter cells. Thus, every nucleated cell you possess, except gametes, contains a complete set of identical genetic instructions for making every structure and performing every function in your body.

How, then, can liver, bone, blood, muscle, and nerve cells look and act so differently from one another? The answer is deceptively simple: Only certain genes are active in a certain type of cell most genes are turned off in any given cell, which leads to specialization for specific jobs. The active genes produce specific proteins that determine the structure and function of that particular cell. Indeed, as cells become specialized for specific jobs, the timing of the activity of specific genes is critical.

But what controls gene activity? The answer to this question is a bit more complex, because gene activity is controlled in several ways. Genes are regulated on several levels simultaneously.

Gene Activity at the Chromosome Level

At the chromosome level, gene activity is affected by the coiling and uncoiling of the DNA. When the DNA is tightly coiled, or condensed, the genes are not expressed. When a particular protein is needed in a cell, the region of the chromosome containing the necessary gene unwinds, allowing transcription to take place. Presumably, the uncoiling allows enzymes responsible for transcription to reach the DNA in that region of the chromosome. Other regions of the chromosome remain tightly coiled and therefore are not expressed. Indeed, the environment may affect which genes are turned on or off (see Environmental Issue essay, Environment and Epigenetics).

Regulating the Transcription of Genes

Some regions of DNA regulate the activity of other regions. As we have seen, a promoter is a specific sequence of DNA that is located adjacent to the gene it regulates. When regulatory proteins called transcription factors bind to a promoter, RNA polymerase can bind to the promoter, which begins transcription of the regulated genes.

Transcription factors can also bind to enhancers, segments of DNA that increase the rate of transcription of certain genes and, therefore, the amount of a specific protein that is produced. Enhancers also specify the timing of expression and a gene's response to external signals and developmental cues that affect gene expression.

You may recall from Chapter 10 that one of the ways certain hormones bring about their effects is by turning on specific genes. Steroid hormones, for instance, bind to receptors within a target cell. The hormone-receptor complex then finds its way to the chromatin in the nucleus and turns on specific genes. For example, one such complex turns on the genes in cells that produce facial hair—explaining why your father may have a beard but your mother probably does not, even though she has the necessary genes to grow one. In this case, the sex hormone testosterone binds to a receptor and turns on hair-producing genes. Facial hair follicle cells of both men and women have the necessary testosterone receptors. However, women usually do not produce enough testosterone to activate the hair-producing genes, so bearded women are rare.

Why do female athletes who inject themselves with testosterone to stimulate muscle development sometimes develop increased facial hair?

Environment and Epigenetics

Your lifestyle may influence the health of your great grandchild. How is this possible? It can occur through epigenetics, which involves a stable alteration in gene expression without changes in DNA sequence. In other words, it regulates how genes are expressed without changing the proteins they encode. We will consider two epigenetic processes: DNA methylation and histone acetylation. These processes alter gene expression by affecting how tightly packaged the DNA molecule is. DNA is packaged with proteins to form chromosomes. DNA methylation (adding a methyl group to the cytosine bases in DNA) turns off the activity of a gene by bringing in proteins that act to compact DNA into a tighter form. On the other hand, histone acetylation makes the DNA less tightly coiled and gene expression easier.

We now know that these processes can be affected by the environment and that the pattern of DNA methylation is dynamic and changes over time. DNA methylation patterns can be affected by environmental factors, cause disease, be transmitted through generations, and, potentially, influence evolution. DNA is sensitive to the environment, so what we eat and the chemicals we are exposed to, including pesticides, tobacco smoke, hormones, and nutrients, may influence our health by affecting our gene expression patterns. For example, maternal nutrition during pregnancy can cause epigenetic changes in gene activity in the fetus that may increase susceptibility to obesity, type-2 diabetes, heart disease, and cancer. The quantity of food consumed during pregnancy alters the offspring's susceptibility to cardiovascular disease. Epigenetics is also thought to play a role in human behavioral disorders, such as autism spectrum disorders (discussed in Chapter 18a), Rett syndrome (a developmental disorder that affects the nervous system), and Fragile-X syndrome (an inherited form of mental impairment). For example, there is some evidence that a gene needed to respond to oxytocin (a hormone important in social bonding) is turned off in some people with autism. As we will see in Chapter 21a, cancer development is controlled by cancer-inhibiting and cancer-promoting genes. If cancer-inhibiting genes are turned off or cancer-promoting genes turned on cancer can result. Changes in the pattern of gene expression are found in cancers of the cervix, prostate, breast, stomach, and colon.

Although DNA methylation patterns are considered to be stable, some studies suggest that methylation can be reversed in adulthood. Foods such as broccoli, onions, and garlic may reduce methylation, allowing genes to be expressed. Researchers are actively looking for drugs that will alter the pattern of methylation and cure cancer.

• Do you think that epigenetics increases or decreases a person's responsibility for their own behavior?

• Researchers may someday develop an “epigenetic diet” that favors positive changes in gene activity. Would you follow that diet? Do you think that pregnant women should be required to follow that diet?

Genetic Engineering

The manipulation of genetic material for human purposes, a practice called genetic engineering, began almost as soon as scientists started to understand the language of DNA. Genetic engineering is part of the broader endeavor of biotechnology, a field in which scientists make controlled use of living cells to perform specific tasks. Genetic engineering has been used to produce pharmaceuticals and hormones, improve diagnosis and treatment of human diseases, increase food production from plants and animals, and gain insight into the growth processes of cells.

The basic idea behind genetic engineering is to put a gene of interest—in other words, one that produces a useful protein or trait—into another piece of DNA to create recombinant DNA, which is DNA combined from two or more sources. The recombinant DNA, carrying the gene of interest, is then placed into a rapidly multiplying cell that quickly produces many copies of the gene. The final harvest may consist of large amounts of the gene product or many copies of the gene itself. Let's take a closer look at the procedure one step at a time.

1. The gene of interest is sliced out of its original organism and spliced into vector DNA. Both the DNA originally containing the gene of interest and the vector DNA, which receives the transferred genes and transports it to a new cell, are cut at specific sequences that are recognized by a restriction enzyme. This is a type of enzyme that makes a staggered cut between specific base pairs in DNA, leaving several unpaired bases on each side of the cut. There are many kinds of restriction enzymes each kind recognizes and cuts a different sequence of DNA. The stretch of unpaired bases produced on each side of the cut is called a sticky end because of its tendency to pair with the single-stranded stretches of complementary base sequences on the ends of other DNA molecules that were cut with the same restriction enzyme (Figure 21.13).

FIGURE 21.13. DNA from different sources can be spliced together using a restriction enzyme to make cuts in the DNA. A restriction enzyme makes a staggered cut at a specific sequence of DNA, leaving a region of unpaired bases on each cut end. The region of singlestranded DNA at the cut end is called a sticky end, because it tends to pair with the complementary sticky end of any other piece of DNA that has been cut with the same restriction enzyme, even if the pieces of DNA came from different sources.

The sticky ends are the secret to splicing the gene of interest and the vector DNA. The sticky ends of DNA from different sources will be complementary and stick together as long as they have been cut with the same restriction enzyme. The initial attachment between sticky ends is temporary, but the ends can be "pasted" together permanently by another enzyme, DNA ligase. The resulting recombinant DNA contains DNA from two sources.

2. The vector is used to transfer the gene of interest to a new host cell. Biological carriers that ferry the recombinant DNA to a host cell are called vectors. A common vector is bacterial plasmids, which are small, circular pieces of self-replicating DNA that exist separately from the bacterial chromosome.2 As previously described, the source DNA (that is, the source of the gene of interest) and the plasmid (vector) DNA are both treated with the same restriction enzyme. Afterward, fragments of source DNA, some of which will contain the gene of interest, will be incorporated into plasmids when their sticky ends join. The recombinant DNA is mixed with bacteria in a test tube. Under the right conditions, some of the bacterial cells will then take up the recombined plasmids (Figure 21.14).

FIGURE 21.14. An overview of genetic engineering using plasmids

Although the basic strategy is usually the same, there are many variations on this theme of transporting a gene into a new host. For instance, the gene of interest is sometimes combined with viral DNA. The viruses are then used as vectors to insert the recombinant DNA into a host cell. Cells other than bacteria, including yeast or animal cells, can also be used as vectors.

3. The recombinant organism containing the gene of interest is identified and isolated from the mixture of recombinants. When plasmids are used as vector DNA, each recombinant plasmid is introduced into a single bacterial cell, and each cell is then grown into a colony. Each colony contains a different recombinant plasmid. The bacteria containing the gene of interest must be identified and isolated.

4. The gene is amplified through bacterial cloning or by use of a polymerase chain reaction. After the colony containing the gene of interest has been identified, researchers usually amplify (that is, replicate) the gene, producing numerous copies. Gene amplification is accomplished using one of two techniques: bacterial cloning or a polymerase chain reaction.

Cloning . Bacteria containing the plasmid with the gene of interest can be grown in huge numbers by cloning. Each bacterium divides many times to form a colony. Thus, each colony constitutes a clone—a group of genetically identical organisms all descended from a single cell. In this case, all the members of the clone carry the same recombinant DNA. Later, the plasmids can be separated from the bacteria, a process that partially purifies the gene of interest. The plasmids then can be taken up by other bacteria that will thus become capable of performing a service deemed useful by humans. Alternatively, the plasmids can be transferred into plants or animal cells— creating transgenic organisms—organisms containing genes from another species.

Polymerase chain reaction (PCR). In PSR (Figure 21.15), the DNA of interest is unzipped, by gentle heating, to break the hydrogen bonds and form single strands. The single strands, which will serve as templates, are then mixed with primers— special short pieces of nucleic acid—one primer with bases complementary to each strand. The primers serve as start tags for DNA replication. Nucleotides and a special heat-resistant DNA polymerase, which promotes DNA replication, are also added to the mixture, which is then cooled to allow base pairing. Through base pairing, a complementary strand forms for each single strand. The procedure is then repeated many times, and each time the number of copies of the DNA of interest is doubled. In this way, billions of copies of the DNA of interest can be produced in a short time.

FIGURE 21.15. The polymerase chain reaction (PCR) rapidly produces a multitude of copies of a single gene or of any desired segment of DNA. PCR amplifies DNA more quickly than does bacterial cloning. It has many uses besides genetic engineering, including DNA fingerprinting.

Genetic engineering involves altering an organism's genes—adding new genes and traits to microbes, plants, or even animals. Do you think we have the right to “play God” and alter life-forms in this way? The U.S. Supreme Court has approved patenting of genetically engineered organisms, first of microbes and now of mammals such as pigs that are genetically modified for use in organ transplant. If you were asked to decide whether it is ethical to patent a new life-form, how would you respond?

Applications of Genetic Engineering

Genetic engineering has been used in two general ways.

• Genetic engineering provides a way to produce large quantities of a particular gene product. The useful gene is transferred to another cell, usually a bacterium or a yeast cell, that can be grown easily in large quantities. The cells are cultured under conditions that cause them to express the gene, after which the gene product is harvested. For example, genetically engineered bacteria have been used to produce large quantities of human growth hormone (Figure 21.16). Treatment with growth hormone allows children with an underactive pituitary gland to grow to nearly normal height.

• Genetic engineering allows a gene for a trait considered useful by humans to be taken from one species and transferred to another species. The transgenic organism then exhibits the desired trait. For example, scientists have endowed salmon with a gene from an eel-like fish. This gene causes the salmon to produce growth hormone year- round (something they do not normally do). As a result, the salmon grow faster than normal.

FIGURE 21.16. Genetic engineering is used to produce large quantities of a desired protein or to create an organism with a desired trait. This boy has an underactive pituitary gland. Its undersecretion of growth hormone would have caused him to be very short, even as an adult. However, growth hormone from genetically engineered bacteria has helped him grow to an almost normal height.

Environmental applications . Genetic engineering also has environmental applications. For example, in sewage treatment plants, genetically engineered microbes lessen the amount of phosphate and nitrate discharged into waterways. Phosphate and nitrate can cause excessive growth of aquatic plants, which could choke waterways and dams, and of algae, which can produce chemicals that are poisonous to fish and livestock. Microorganisms are also being genetically engineered to modify or destroy chemical wastes or contaminants so that they are no longer harmful to the environment. For instance, oil-eating microbes that can withstand the high salt concentrations and low temperatures of the oceans have proven useful in cleaning up after marine oil spills.

Livestock . Genetic engineering has also been used on livestock. Genetically engineered vaccines have been created to protect piglets against a form of dysentery called scours, sheep against foot rot and measles, and chickens against bursal disease (a viral disease that is often fatal). Genetically engineered bacteria produce bovine somatotropin (BST), a hormone naturally produced by a cow's pituitary gland that enhances milk production. Injections of BST can boost milk production by nearly 25%.

Transgenic animals have been created by injecting a fertilized egg with the gene of interest in a petri dish. The goals of creating transgenic animals include making animals with leaner meat, sheep with softer wool, cows that produce more milk, and animals that mature more quickly.

Pharmaceuticals . Genes have been put into a variety of cells, ranging from microbes to mammals, to produce proteins for treating allergies, cancer, heart attacks, blood disorders, autoimmune disease, and infections.

Genetically engineered bacteria have also been used to create vaccines for humans. You may recall from Chapter 13 that a vaccine typically uses an inactivated bacterium or virus to stimulate the body's immune response to the active form of the organism. The idea is that the body will learn to recognize proteins on the surface of the infectious organism and mount defenses against any organism bearing those proteins. Because the organism used in the vaccine was rendered harmless, the vaccine cannot trigger an infection. Scientists produce genetically engineered vaccines by putting the gene that codes for the surface protein of the infectious organism into bacteria. The bacteria then produce large quantities of that protein, which can be purified and used as a vaccine. The vaccine cannot cause infection, because only the surface protein is used instead of the infectious organism itself.

Plants have also been used to produce therapeutic proteins. Engineered bananas that produce an altered form of the hepatitis B virus surface protein are being developed as an edible vaccine against the liver disease hepatitis B. Someday, you may eat a banana in order to be vaccinated, instead of receiving an injection against hepatitis B. Plants are also being engineered to produce "plantibodies," antibodies made by plants. For example, soybeans are being cultivated that contain human antibodies to the herpes simplex virus that causes genital herpes. A human gene for an antibody that binds to tumor cells has been transplanted into corn. The antibodies can then deliver radioisotopes to cancer cells, selectively killing them.

Pharming is a word that comes from the combination of the words farming and pharmaceuticals. In gene pharming, transgenic animals are created that produce a protein with medicinal value in their milk, eggs, or blood. The protein is then collected and purified for use as a pharmaceutical. When the pharm animal is a mammal, the gene is expressed in mammary glands. The desired protein is then extracted and purified from milk (Figure 21.17). For example, the gene for the protein alpha-1-antitrypsin (AAT) has been inserted into sheep that then secrete AAT in their milk. People with an inherited, potentially fatal form of emphysema (a lung disease) take AAT as a drug. It is also being tested as a drug to prevent lung damage in people with cystic fibrosis. The first drug made from the milk of a transgenetic goat was an anticlotting drug called ATryn. It is given to people with a blood-clotting deficiency when they must undergo surgery. Researchers have also created a transgenic goat to produce milk containing lysozyme, an antibacterial agent. Lysozyme can be used to treat intestinal infections that kill millions of children in underdeveloped countries.

FIGURE 21.17. The procedure for creating a transgenic animal that will produce a useful protein in its milk

Raw materials . Genetic engineering is also used to produce useful new raw materials. For example, spider silk is five times stronger than steel and still lightweight. Attempts have been made to farm spiders for their silk, but spiders are too aggressive to live close together. Goats have been genetically engineered to possess the gene for spider silk and secrete spider silk proteins in their milk (Figure 21.18). More recently, E. coli bacteria have been genetical engineered to produce spider silk proteins. The spider silk proteins can be spun into a fine thread, which could be used for bulletproof clothing, thinner surgical thread, and stronger yet lighter racing cars and aircraft.

FIGURE 21.18. This transgenic goat has the gene for making spider silk, one of the strongest substances known. The spider silk protein can be extracted from the goat’s milk and spun into threads that can be used for products in which strength and light weight are important qualities.

Agriculture . Most of us experience some of the results of genetic engineering at our dinner tables (see Health Issue Essay, Genetically Modified Food.) The most common traits that have been genetically engineered into crops are resistance to pests and resistance to herbicides. Scientists also have developed two virus-resistant strains of papaya and distributed them to papaya growers in Hawaii, saving the industry from ruin. In addition, different strains of rice have been genetically engineered to resist disease-causing bacteria and to withstand flooding of the paddy. Other plants have been genetically engineered to be more nutritious. For example, golden rice is a strain of rice that has been genetically engineered to produce high levels of beta-carotene, a precursor of vitamin A, which is in short supply in certain parts of the world. More than 100 million children worldwide suffer from vitamin A deficiency, and 500,000 of them go blind every year because of that deficiency. Although golden rice cannot supply a complete recommended daily dose of vitamin A, the amount it contains could be helpful to a person whose diet is very low in vitamin A. Other crops have also been created that grow faster, produce greater yields, and have longer shelf lives.

The problems associated with many genetic diseases arise because a mutant gene fails to produce a normal protein product. The goal of gene therapy is to cure genetic diseases by putting normal, functional genes into the body cells that were affected by the mutant gene. The functional gene would then produce the needed protein.

Methods of delivering a healthy gene . One way a healthy gene can be transferred to a target cell is by means of viruses. Viruses generally attack only one type of cell. For instance, an adenovirus, which causes the common cold, typically attacks cells of the respiratory system. A virus consists largely of genetic material, commonly DNA, surrounded by a protein coat (see Chapter 13a). Once inside a cell, the viral DNA uses the cell's metabolic machinery to produce viral proteins. If a healthy gene is spliced into the DNA of a virus that has first been rendered harmless, the virus will deliver the healthy gene to the host cell and cause the desired gene product to be produced (Figure 21.19).

Another type of virus used in gene therapy is a retrovirus, a virus whose genetic information is stored as RNA rather than DNA. Once inside the target cell, a retrovirus rewrites its genetic information as double-stranded DNA and inserts the viral DNA into a chromosome of the target cell.

FIGURE 21.19. Gene therapy using a virus. In gene therapy, a healthy gene is introduced into a patient who has a genetic disease caused by a faulty gene.

Genetically Modified Food

From dinner tables to diplomatic circles, people are discussing genetically modified (GM) food. This relatively recent interest is somewhat ironic, considering that people in the United States have been eating GM food since the mid-1990s. More than 70% of processed foods sold in the United States contain genetically modified ingredients. Yet many people vehemently object to GM food.

Why is something as common as GM food controversial? The concerns about it can generally be divided into three categories: health issues, social issues, and environmental issues. Let's explore these categories one at a time.

A panel of the National Academy of Sciences (NAS) has issued a report saying that genetically engineered crops do not pose health risks that cannot also be caused by crops created by conventional breeding. However, because genetic engineering could produce unintended harmful changes in food, the NAS panel recommends scrutiny of GM foods before they can be marketed. Currently, the U.S. Department of Agriculture, the Food and Drug Administration (FDA), and the Environmental Protection Agency regulate genetically modified foods. The NAS panel concluded that the GM foods already on the market are safe.

· The larger salmon in the back of the photo has been genetically modified to grow faster than the normal salmon in front.

A common safety concern is that GM foods may contain allergens (substances that cause allergies). After a protein is produced, the cell modifies it in various ways. The protein may be modified in the genetically modified plant differently from the way it would in an unmodified cell, and the modification could produce an allergen. The likelihood that this may occur is reduced by rigorous testing. Most known allergens share certain properties. They are proteins, relatively small molecules, and resistant to heat, acid, and digestion in the stomach. If a protein produced by a GM plant has any of the properties typical of an allergen or is structurally similar to a known allergen, the FDA considers it to be a potential allergen and requires that the protein undergo additional allergy testing.

Bacterial resistance to antibiotics is likely to be a major threat to public health in this decade (see Chapter 13a). When bacteria are resistant to an antibiotic, the drug will not kill them and thus will no longer cure the human disease for which they are the cause. Some people worry about the scientific practice of putting genes for resistance to an antibiotic into GM crops as markers to identify the plants with the modified genes. Plant seedlings thought to be genetically modified are grown in the laboratory in the presence of an antibiotic. Only those seedlings with the gene for resistance will survive. Because of the way they were engineered, the surviving plants also contain the “useful” gene.

What worries some people is that the genes for resistance to antibiotics could be transferred to bacteria, making the bacteria resistant to antibiotics. The receiving bacteria might be those that normally live in the human digestive system, or they might be bacteria ingested with food. It is not known whether genes can be transferred from a plant to a bacterium. However, it is known that bacteria can easily and quickly transfer genes for antibiotic resistance to one another. Thus, a harmless bacterium in the gut could transfer the gene for antibiotic resistance to a disease-causing bacterium.

The transfer of antibiotic-resistance genes from GM plants to bacteria could have serious consequences. For this reason, antibiotic- resistance marker genes are being phased out in favor of other marker genes, such as a green fluorescent protein. Scientists also have developed a way to inactivate the antibiotic-resistance gene if it were to be transferred to bacteria.

Proponents of GM foods argue that herbicide- resistant and pesticide-resistant crops reduce the need for spraying with herbicides and pesticides. So far, experience has shown that the validity of this argument depends on the crop. Pest-resistant cotton has substantially reduced the use of pesticides, but pest-resistant corn probably has not. Farmers who grow herbicide- resistant crops still spray with herbicides, but they change the type of herbicide they use to a type that is less harmful to animals.

Unfortunately, engineered crops containing insecticides could have undesirable effects on insects. Genetically engineering insecticides into plants could hasten the development of insect resistance to that insecticide, making the insecticide ineffective—not just for the genetically modified crop, but for all crops.

Another concern is that genetically modified organisms could harm other organisms. One example is that pollen from pest-resistant corn has been shown to harm monarch butterfly caterpillars. Fortunately, monarch caterpillars rarely encounter enough pollen to be harmed, and most of the pest-resistant corn grown in the United States today does not produce pollen that is harmful to monarch butterflies. A second example of a genetically modified organism that has the potential to harm other organisms is the salmon that produce more growth hormone and grow several times faster than their wild relatives do. These salmon are grown on fish farms. If the FDA grants approval, the gene-altered salmon could dramatically cut costs for fish farmers and consumers. However, when the genetically modified salmon are grown in tanks with ordinary salmon and food is scarce, the genetically modified salmon eat most of the food and some of their ordinary companions. What would happen if the genetically modified salmon escaped from their pens on the fish farm? They might mate with the wild salmon and create less healthy offspring or outcompete wild salmon for food, which could eventually cause the extinction of the wild salmon. Escape is possible. During the past few years, hundreds of thousands of fish have escaped from fish farms when floating pens were ripped apart by storms or sea lions. To minimize the risk that genetically modified salmon could destroy the population of wild salmon, scientists plan to breed the fish inland, sterilize the offspring, and ship only sterile fish to coastal pens. The sterilization procedure is effective in small batches of fish, but it is not known whether the procedure is completely effective in large batches of fish.

Critics of GM foods also fear that crops genetically engineered to resist herbicides could become "super weeds" that could not be controlled with existing chemicals. In North Dakota, GM canola plants that are resistant to herbicides are growing along roadsides. Canola can hybridize with at least two wild weeds. The two original strains of GM canola were each resistant to a different herbicide. As a result of cross-pollination, some of the canola plants found in the wild are resistant to both herbicides, which suggests that the GM traits are stable in the wild and are evolving.

Proponents of GM food claim that GM food can assist in the battle against world hunger. We have seen that genetic engineering can produce crops that resist pests and disease. It can also produce crops with greater yields and crops that will grow in spite of drought, depleted soil, or excess salt, aluminum, or iron. Foods can also be genetically modified to contain higher amounts of specific nutrients.

Critics of using GM food to battle world hunger argue that the problem of hunger has nothing to do with an inability to produce enough food. The problem, they say, is a social one of distributing food so that it is available to the people who need it.

Some developing countries are resisting the use of GM seeds. Part of the resistance stems from lingering health concerns. However, many farmers in developing countries also object to GM seeds because the GM plants do not seed themselves. The need to buy seeds each year places a financial burden on poor farmers.

• Do you consider GM food to be a blessing or a danger to the world? What are your reasons?

• If a genetically modified organism that was created for food begins to cause environmental problems, who should be held accountable?

• Do you think that foods containing genetically modified components should be labeled as such?

Gene therapy results . More than 4000 human diseases have been traced to defects in single genes. Although the Food and Drug Administration has not yet approved a gene therapy for any of these conditions, hundreds of clinical trials of gene therapies are currently under way, including trials of possible therapies for cystic fibrosis and cancer (discussed in Chapter 21a).

The first condition to be treated experimentally with gene therapy was a disorder referred to as severe combined immunodeficiency disease (SCID). The immune system of children with SCID is nonfunctional, leaving them vulnerable to infections. The cause of the problem is a mutant gene that prevents the production of an enzyme called adenosine deaminase (ADA). Without ADA, white blood cells never mature—they die while still developing in the bone marrow. The first gene therapy trial began in 1990, when white blood cells of a 4-year-old SCID patient, Ashanthi DeSilva, were genetically engineered to carry the ADA gene and then returned to her tiny body. Her own gene-altered white blood cells began producing ADA, and her body defense mechanisms were strengthened. Ashanthi's life began to change. She was not ill as often as she had been before. She could play with other children. However, the life span of white blood cells is measured in weeks, and when the number of gene-altered cells declined, new gene-altered cells had to be infused. Ashanthi is now in her twenties and has a reasonably healthy immune system. However, she still needs repeated treatments.

French scientists believe they have cured 10 children with X-SCID by using gene therapy (Figure 21.20). X-SCID is a severe combined immunodeficiency syndrome caused by a mutant gene on the X chromosome. It is still too soon to know for certain whether all 10 children will require treatment in the future. However, four children in the French studies who had shown improvement in symptoms of X-SCID developed leukemia from the therapy.

FIGURE 21.20. Rhys Evans is the first person to be cured of X-linked severe combined immunodeficiency disease (X-SCID) by gene therapy. The immune system of a person with X-SCID is nonfunctional. Rhys Evans’s immune system was strengthened, and he can now go to public places without fearing contact with people who might carry germs. He can play with other children.

A form of muscular dystrophy has recently been treated with gene therapy. Researchers packed a virus with the normal form of the protein that is deficient in this form of muscular dystrophy. The virus was then injected directly into the muscles. The level of the protein and the gene expression of the protein remained high for several months, restoring some muscle function to the patients.

Without treatment, children with X-SCID die at a young age. The gene therapy treatment for X-SCID that caused leukemia in four French boys does appear to have cured this deadly disorder in other patients. If you had a child with X-SCID, would you want him to have this gene therapy treatment? Why or why not? What factors would you consider in making your decision?

A genome is the entire set of genes carried by one member of a species—in our case, one person. Genomics is the study of entire genomes and the interactions of the genes with one another and the environment.

One goal of genomics is to determine the location and sequences of genes. Researchers have developed supercomputers that automatically sequence (determine the order of bases in) DNA. The supercomputers were put to use on a massive scale in the Human Genome Project, a worldwide research effort, completed in 2003, to sequence the human genome. As a result, we now have some idea of the locations of genes along all 23 pairs of human chromosomes and the sequence of the estimated 3 billion base pairs that make up those chromosomes. Although the exact number of human genes is still not known for certain, scientists now estimate that the human genome consists of 20,000 to 25,000 genes, not 100,000 as originally thought. One reason for the smaller number of actual genes is that many gene families have related or redundant functions and therefore are able to share certain genes, so that fewer are needed to carry out all the body's functions. A second reason is that many genes are now known to code for parts of more than one protein.

In addition, researchers have identified and mapped to specific locations on specific chromosomes genes for more than 1400 genetic diseases. It is hoped that this information will give scientists a greater ability to diagnose, analyze, and eventually treat many of the 4000 diseases of humans that have a known genetic basis. Researchers have already cloned the genes responsible for many genetic diseases, including Duchenne muscular dystrophy, retinoblastoma, cystic fibrosis, and neurofibromatosis. These isolated genes can now be used to test for the presence of the same disease-causing genes in specific individuals. As we saw in Chapter 20, some gene tests can be used to identify people who are carriers for certain genetic diseases such as cystic fibrosis, allowing families to make choices based on known probabilities of bearing an affected child similar tests can be used for prenatal diagnosis and for diagnosis before symptoms of the disease begin. After a disease-related gene has been identified, scientists can study it to learn more about the protein it codes for and perhaps discover ways to correct the problem.

We have also learned from the Human Genome Project that humans are identical in 99.9% of the sequences of their genes. As scientists gain greater understanding of the 0.1% of DNA that differs from person to person, they expect to learn more about why some people develop heart disease, cancer, or Alzheimer's disease and others do not.

A second goal of genomics is to understand the mechanisms that control gene expression. More than 95% of human DNA does not code for protein however, some of these noncoding DNA sequences function as regulatory regions that determine when, where, and how much of certain proteins are produced. Because gene activity plays a role in many diseases, the study of how these regions turn genes on or off may lead to advances in diagnosis and treatment.

One of the tools researchers use in this effort is the microarray, which consists of thousands of DNA sequences stamped onto a single glass slide called a DNA chip. Researchers use microarrays to monitor large numbers of DNA segments to discover which genes are active and which are turned off under different conditions, such as in different tissue types, different stages of development, or in health and disease. For example, they may use microarrays to identify genes that are active in cancerous cells but not in healthy cells (Figure 21.21). Presumably, the genes that are active in cancerous cells play a role in the development of cancer.

FIGURE 21.21. A comparison of microarrays showing the pattern of gene activity in prostate cancer and in normal tissue. The genes that are active in tissue with prostate cancer but not in normal tissue probably play a role in the development of cancer.

Besides identifying gene activity in health and disease, microarray analysis is useful in identifying genetic variation in the members of a population. Some of these genetic differences are in the form of single-nucleotide polymorphisms (SNPs, or snips). These are DNA sequences that can vary by one nucleotide from person to person, and the differences in their protein products are thought to influence how we respond to stress and diseases, among other things. As researchers learn more about SNPs, they may be able to develop treatments tailored to the genetic makeup of each individual. These are the kinds of discoveries that can open the door for gene therapy. While individualized gene therapy may be useful in the future, we already use identification of individual differences in DNA on a regular basis in DNA fingerprinting (see the Ethical Issue essay, Forensic Science, DNA, and Personal Privacy).

Some people worry that once we know the location and function of every gene and have perfected the techniques of gene therapy, we will no longer limit gene manipulation to repairing faulty genes but will begin to modify genes to enhance human abilities. Should people be permitted to design their babies by choosing genes that they consider superior? What do you think? Where should the line be drawn? Who should draw that line? Who should decide which genes are “good?”

When SNPs are located near one another on a chromosome, they tend to be inherited together. A group of SNPs in a region of a chromosome is called a haplotype. The International HapMap Project is a scientific consortium whose purpose is to describe genetic variation between populations. Researchers collaborating on this project compare haplotype frequencies in groups of people who have a certain disease to those of a group without the disease, hoping to identify genes associated with the disease.

Comparison of Genomes of Different Species

The DNA of certain widely studied organisms, including the mouse, the fruit fly, a roundworm, yeast, slime mold, and the honeybee have also been mapped. From these genomes, geneticists hope to gain some insight into basic biology, including basic principles of the organization of genes within the genome, gene regulation, and molecular evolution. Humans share many genes with other organisms. For example, we share 50% of our genes with the fruit fly and 90% of our genes with the mouse. These genetic similarities are evidence of our common evolutionary past. The genes and genetic mechanisms we share with other organisms are likely to be important in determining body form as well as influencing development and aging.

In Chapter 19, we learned about the cell cycle. In Chapters 20 and 21, we learned about genes, their inheritance, and their regulation, and we also considered how mutations affect gene functions. In Chapter 21a, we use this information to understand cancer, a family of diseases in which mutations in genes that regulate the cell cycle cause a loss of control over cell division.

Forensic Science, DNA, and Personal Privacy

So-called DNA fingerprints, like the more conventional prints left by fingers, can help identify the individuals they belong to out of a large population. DNA fingerprinting refers to techniques of identifying individuals on the basis of unique features of their DNA. DNA fingerprints are possible because many regions of DNA are composed of small, specific sequences of DNA that are repeated many times. Most commonly used are repeated units of 1 to 5 bases, which are called short tandem repeats (STRs). The number of times these sequences are repeated varies considerably from person to person, from a few to 100 repeats. Because of these differences, the segments can be used to match a sample of DNA to the person whose cells produced the sample.

The first step in preparing a DNA fingerprint is to extract DNA from a tissue sample. The type of tissue does not matter, and one type of tissue can be successfully compared to another type. Commonly used sources include blood, semen, skin, and hair follicles, because they are not too painful to remove, they are readily available, or they are left at a crime scene.

First, the amount of DNA is greatly increased using PCR, as described in Figure 21.15. The primers used are sequence specific for the regions on either side of the repeating region. This produces many copies of the repeating region, which are then analyzed to determine the number of repeats present.

FIGURE 21.A. The pattern of banding in a DNA fingerprint is determined by the sequence of bases in a person’s DNA and is, therefore, unique to each person. A match between DNA fingerprints can identify the source of a tissue sample from a crime scene with a high degree of certainty. Which suspect’s DNA fingerprint matches this specimen found at a crime scene?

The FBI uses 13 STRs as a core set for forensic analysis. The resulting DNA fingerprint is unique to the person who produced the DNA. Moreover, any DNA sample taken from the same person would always be identical. But the fingerprint profiles resulting from the DNA of different people are always different (except perhaps for identical siblings), because the number and sizes of the fragments are determined by the unique sequence of bases in each person’s DNA.

DNA fingerprinting has many applications, but the most familiar is probably its use in crime investigations. In these cases the DNA fingerprint is usually created from a sample of tissue, such as blood or hair follicles, collected at the crime scene. A fingerprint can be produced from tissue left at the scene years before. This fingerprint is then compared with the DNA fingerprints of various suspects. A match reveals, with a high degree of certainty, the person who was the source of the sample from the crime scene (Figure 21.A).

Of course, the degree of certainty of a match between DNA fingerprints depends on how carefully the analysis was done. Because DNA fingerprints are being used as evidence in an increasing number of court cases each year, it is important that national standards be set to ensure the reliability of these molecular witnesses. It is generally easier to declare with certainty that two DNA fingerprints do not match than it is to be sure that they do. In the United States more than 200 convicts have been found innocent through DNA testing during the last decade.

Has DNA testing gone too far? All states in the United States collect DNA samples from people convicted of sex crimes and murder. Several other states also collect DNA samples from people convicted of other felonies, such as robbery. About 30 states collect DNA samples from people accused of misdemeanors, including loitering, shoplifting, or vandalism. In many cases, the DNA is stored in a database, even if the person is found innocent of the crime. People who are simply cooperating with the investigation may also provide DNA samples. These, too, are added to a national database.

• Do you think everyone arrested of a crime should have the right to DNA fingerprinting to prove his or her innocence? If so, who should pay for the process?

• Is the creation of a national database of DNA fingerprints an invasion of privacy? Is it any more so than a database of actual fingerprints or mug shots?

• Under what conditions do you think DNA samples should be obtained?

Highlighting the Concepts

• DNA consists of two strands of nucleotides linked by hydrogen bonds and twisted together to form a double helix. Each nucleotide consists of a phosphate, a sugar called deoxyribose, and one of four nitrogenous bases: adenine, thymine, cytosine, or guanine. The sugar and phosphate components of the nucleotides alternate along the two sides of the molecule. Pairs of bases meet and form hydrogen bonds in the interior of the double helix these pairs resemble the rungs on a ladder.

• According to the rules of complementary base pairing, adenine binds only with thymine, and cytosine binds only with guanine.

Replication of DNA (pp. 435-436)

• DNA replication is semiconservative. Each new double-stranded DNA molecule consists of one old and one new strand. The enzyme DNA polymerase "unzips" the two strands of a molecule (the parent molecule), allowing each strand to serve as a template for the formation of a new strand. Complementary base pairing ensures the accuracy of replication.

Gene Expression (pp. 436-441)

• Genetic information is transcribed from DNA to RNA and then translated into a protein.

• Transcription is the synthesis of RNA by means of base pairing on a DNA template. RNA differs from DNA in that the sugar ribose replaces deoxyribose, and the base uracil replaces thymine. Most RNA is single stranded.

• Messenger RNA (mRNA) carries the DNA genetic message to the cytoplasm, where it is translated into protein. The genetic code is read in sequences of three RNA nucleotides each triplet is called a codon. Each of the 64 codons specifies a particular amino acid or indicates the point where translation should start or stop.

• Transfer RNA (tRNA) interprets the genetic code. At one end of the tRNA molecule is a sequence of three nucleotides called the anticodon that pairs with a codon on mRNA in accordance with basepairing rules. The tail of the tRNA binds to a specific amino acid.

• Each of the two subunits of a ribosome consists of ribosomal RNA (rRNA) and protein. A ribosome brings tRNA and mRNA together for protein synthesis.

• Translation of the genetic code into protein begins when the two ribosomal subunits and an mRNA assemble, with the mRNA sitting in a groove between the ribosome's two subunits. The mRNA attaches to the ribosome at the mRNA's start codon. Then the ribosome slides along the mRNA molecule, reading one codon at a time. Molecules of tRNA ferry amino acids to the mRNA and add them to a growing protein chain. Translation stops when a stop codon is encountered. The protein chain then separates from the ribosome.

• A point mutation is a change in one or a few nucleotides in the sequence of a DNA molecule. When one nucleotide is mistakenly substituted for another, the function of the resulting protein may or may not affect the function of the protein. The insertion or deletion of a nucleotide always changes the resulting protein.

Regulating Gene Activity (p. 442)

• Gene activity is regulated at several levels. Usually, most of the DNA is folded and coiled. For a gene to be active, the region of DNA in which it is located must be uncoiled. Gene activity can be affected by other segments of DNA. Regions of DNA called enhancers can increase the amount of RNA produced. Chemical signals such as regulatory proteins or hormones can also affect gene activity.

Genetic Engineering (pp. 442-450)

• Genetic engineering is the purposeful manipulation of genetic material by humans. It can be used to produce large quantities of a particular gene product or to transfer a desirable genetic trait from one species to another or to another member of the same species.

• Genetic engineering uses restriction enzymes to cut the source DNA, which contains the gene of interest, and the vector DNA at specific places, creating sticky ends composed of unpaired complementary bases that allow the cut segments to recombine. The recombinant vector is then used to transfer the recombinant DNA to a host cell. Common vectors include bacterial plasmids and viruses. The host cell is often a type of cell that reproduces rapidly, such as a bacterium or yeast cell. Each time a host cell divides, both daughter cells receive a copy of the gene of interest. Introducing a gene from one species into a different species results in a transgenic plant or animal.

• Genetic engineering has had many applications in plant and animal agriculture, environmental science, and medicine.

• In gene therapy, a healthy form of a gene is introduced into body cells to correct problems caused by a defective gene.

• A genome consists of all the genes in a single organism.

Genomics is the study of genomes and the interaction between genes and the environment. It is now believed that the human genome consists of 20,000 to 25,000 genes.

• Scientists use microarrays to analyze gene activity under different conditions. It uses DNA chips—glass slides with thousands of DNA segments stamped on them. The information may be helpful in treating genetic diseases. Microarray analysis also allows scientists to discover small differences in the gene sequences of people. Scientists hope to use this information to develop individualized treatments.

• Large portions of our DNA are the same as the DNA in other organisms. The closer the evolutionary relationship, the greater portion of DNA we have in common.

1. Describe the structure of DNA. p. 434

2. Explain why complementary base pairing is crucial to exact replication of DNA. p. 435

3. Why is DNA replication described as semiconservative? p. 436

4. Explain the roles of transcription and translation in converting the DNA message to a protein. pp. 436-437

5. In what ways does RNA differ from DNA? pp. 436-437

6. What roles do mRNA, tRNA, and rRNA play in the synthesis of protein? p. 437

7. Define codon. What role do codons play in protein synthesis? p. 438

8. What is the role of an anticodon? pp. 438-439

9. Describe the events that occur during the initiation of protein synthesis, the elongation of the protein chain, and the termination of synthesis. pp. 439-441

10. Why does a deletion have such a major effect on a cell? pp. 441-442

11. How is gene activity regulated? p. 442

12. Define genetic engineering. Explain the roles of restriction enzymes and vectors in genetic engineering. pp. 442-445

13. Describe some of the ways in which genetic engineering has been used in farming and medicine. pp. 445-450

14. What is gene therapy? pp. 448-450

15. The complementary base for thymine is

16. Although the amount of any particular base in DNA will vary among individuals, the amount of guanine will always equal the amount of

b. mRNA complementary to a template strand of DNA.

c. tRNA complementary to mRNA.

19. The anticodon is located on a molecule of _____.

20. In RNA, the nucleotide _____ binds with adenine.

21. In genetic engineering, the staggered cuts in DNA that allow genes to be spliced together are made by ____.

Use the genetic code in Figure 21.5 (p. 438) to answer questions 1 and 2.

1. What would be the amino acid sequence in the polypeptide resulting from a strand of mRNA with the following base sequence?

2. The following are base sequences in four mRNA strands: one normal and three with mutations. (Keep in mind that translation begins with a start codon and ends with a stop codon.) Which of the mutated sequences is likely to have the most severe effects? Why? Which would have the least severe effects? Why?

Normal mRNA: AUG ACA UAU GAG ACG ACU

Mutation 1: AUG ACC UAC GAA ACG ACC

Mutation 2: AUG ACU UAA GAG ACG ACA

Mutation 3: AUG ACG UAU GAG ACG ACG

3. You are a crime scene investigator testifying at a murder trial. The DNA fingerprints shown on the right are those of a bloodstain at the murder scene (not the victim's blood) and those of seven suspects (numbered 1 through 7). Which suspect's blood matches the bloodstain from the crime scene?

Becoming Information Literate

Describe some of the possible medical uses of information gained from the Human Genome Project. What ethical issues are raised by the project? Use at least three reliable sources (journals, books, websites) to answer these questions. Explain why you chose those sources.

1 We will develop this concept as the chapter proceeds. Some genes code for a polypeptide that is only part of a functional protein. A gene can also code for RNA that forms part of a ribosome or that transports amino acids during protein synthesis.

2 Plasmids seem to have evolved as a means of moving genes between bacteria. A plasmid can replicate itself and pass, with its genes, into another bacterium.

If you are the copyright holder of any material contained on our site and intend to remove it, please contact our site administrator for approval.


GENMARK: Parallel gene recognition for both DNA strands ☆

The problem of predicting gene locations in newly sequenced DNA is well known but still far from being successfully resolved. A novel approach to the problem based on the frame dependent (non-homogeneous) Markov chain models of protein-coding regions was previously suggested. This approach is, apparently, one of the most powerful “search by content” methods. The initial idea of the method combines the specific Markov models of coding and non-coding region together with Bayes' decision making function and allows easy generalization for employing of higher order Markov chain models. Another generalization which is described in this article allows the analysis of both DNA strands simultaneously. Currently known gene searching methods perform the analysis of the two DNA strands in turn, one after another. In doing this all the known methods fail in the sense that they generate false (artifactual) prediction signals for the given strand when the real coding region is located on the complementary DNA strand. This common drawback is avoided by employing the Bayesian algorithm which uses an additional non-homogeneous Markov chain model of the “shadow” of the coding region—the sequence which is complementary to the protein-coding sequence.

The preliminary version of this work was presented during the Second International Workshop of Open Problems in Computational Molecular Biology, Telluride Summer Research Center, Telluride, Colo., 19 July–2 August, 1992.


Samacheer Kalvi 12th Bio Zoology Molecular Genetics Text Book Back Questions and Answers

Question 1.
Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 2.
DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer:
(c) Nucleotide containing sugars, nitrogen bases and phosphates

Question 3.
A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer:
(b) Transcription

Question 4.
The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer:
(d) 3.1 billion

Question 5.
E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer:
(d) One low and one intermediate density band

Question 6.
What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5’ end of the molecules
(b) DNA ligase works only in the 3’ → 5’ direction
(c) DNA polymerase can join new nucleotides only to the 3 ’ end of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5’ end
Answer:
(d) Helicases and single-strand binding proteins that work at the 5’ end

Question 7.
Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 8.
Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semi – conservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer:
(b) Replication occurs as each base is paired with another exactly like it

Question 9.
Which of the following statements is not true about DNA replication in eukaryotes?
(a)) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer:
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.

Question 10.
The first codon to be deciphered was which codes for
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer:
(c) UUU, Phenylalanine

Question 11.
Meselson and Stahl’s experiment proved __________
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer:
(d) Semi-conservative nature of DNA replication

Question 12.
Ribosomes are composed of two subunits the smaller subunit of a ribosome has a binding site for and the larger subunit has two binding sites for two
Answer:
mRNA, tRNA

Question 13.
Anoperonisa:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer:
(d) Gene that switched other genes on or off

Question 14.
When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer:
(d) Both (a) and (b) are correct

Question 15.
Give reasons: Genetic code is ‘universal’.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.

Question 16.
Name the parts marked ‘A’ and ‘B’ in the given transcription unit:
Answer:

Question 17.
Differentiate – Leading strand and lagging strand
Answer:

  1. DNA polymerase I Involved DNA repair mechanism
  2. DNA polymerase II Involved DNA repair mechanism
  3. DNA polymerase III Involved in DNA replicaton

Question 18.
Differentiate – Template strand and coding strand.
Answer:

  1. Template Strand: During replication, DNA strand having the polarity 3’ → 5’ act as template strand.
  2. Coding Strand: During replication, DNA strand having the polarity 5’ → 3’ act as coding strand.

Question 19.
Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.
Answer:
Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotide polymorphism – pronounced as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

Question 20.
State any three goals of the human genome project.
Answer:

  1. Identify all the genes (approximately 30000) in human DNA.
  2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
  3. To store this information in databases.

Question 21.
In E.coli, three enzymes 0- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 22.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional, activity of the structural gene.

  1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
  2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
  3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 23.
A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer:
One of the enzyme synthesized by lac operon is permease which is involved in the transport of lactose into the cells. If the lac operon gets inactivated, permease is not synthesized hence lactose cannot enter the cell. Lactose acts as a inducer, binding to the repressor protein and switch on the operator to initiate gene expression.

Question 24.
Why the human genome project is called a mega project?
Answer:
The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 >109 bp. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

Question 25.
From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer:
Inference of Watson and Crick on DNA replication: They concluded that each of the DNA strand in a helix act as template during DNA replication leading to formation of new daughter DNA molecules, which are complementary to parental strand, (i.e., Semi-conservative method of replication) Inference on coding capability: During transcription, the genetic information in the DNA strand is coded to mRNA as complementary bases, (except for uracil in place of thymine in RNA) Inference on mutation: Any changes in the nucleotide sequence of DNA leads to corresponding alteration in aminoacid sequence of specific protein thus confirming the validity of genetic code.

Question 26.
Why tRNA is called an adapter molecule?
Answer:
The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.

Question 27.
What are the three structural differences between RNA and DNA?
Answer:
DNA:

  1. Sugar is deoxyribose sugar.
  2. Double stranded structure.
  3. Nitrogen bases are Adenine, Guanine, Cytosine and Thymine.
  1. Sugar is ribose sugar.
  2. Single stranded molecule.
  3. Nitrogen bases are Adenine, Guanine, Cytosine and Uracil.

Question 28.
Name the anticodon required to recognize the following codons:
AAU, CGA, UAU, and GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 29.
(a) Identify the figure given below
(b) Redraw the structure as a replicating fork and label the parts
(c) Write the source of energy for this replication and name the enzyme involved in this process.
(d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
(a) Replication fork
(b)

(c) Deoxy nucleotide, triphosphate acts as a energy source for replication. DNA polymerase is used for replication
(d) mRNA contacting information for protein synthesis will developed from DNA strand having polariy 5’ → 3’

Question 30.
If the coding sequence in a transcription unit is written as follows:
5’ TGCATGCATGCATGCATGCATGCATGC 3’
Write down the sequence of mRNA.
Answer:
mRNA sequence is 3’ACGUACGUACGUUCGUACGUACGUACG5’

Question 31.
How is the two stage process of protein synthesis advantageous?
Answer:
The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals. Introns would have arosen before or after the evolution of eukaryotic gene.

If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific ‘target sites’ acting like mobile transposon-like elements (that mediate transfer of genes between organisms – Horizontal Gene Transfer – HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on Earth.

Question 32.
Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer:
Generally proteins contain sulphur but not phosphorous and nucleic acid (DNA) contains , phosphorous but not sulphur. Hence Hershey – Chase used radioactive isotopes of sulphur ( 3 5 S) and phosphorus ( 32 P) to keep separate track of viral protein and nucleic acid in culture medium. The expected result cannot be achieved, if radioactive carbon and nitrogen is used, since these molecules are present in both DNA and proteins.

Question 33.
Explain the formation of a nucleosome.
Answer:
Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere.

The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.

Question 34.
It is established that RNA is the first genetic material. Justify giving reasons.
Answer:
Three molecular biologists in the early 1980’s (Leslie Orgel, Francis Brick and Carl Woese) independently proposed the ‘RNA world’ as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term ‘RNA world’ first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic on Earth. There is now enough evidence to suggest that essential life processes (such as metabolism, translation and splicing etc.,) evolved around RNA. RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalysed by RNA. This catalytic RNA is known as ribozyme. But, RNA being a catalyst was reactive and hence unstable.

This led to evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double stranded molecule having complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as the genetic material. Andrew Fire and Craig Mellow (recipients of Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

Samacheer Kalvi 12th Bio Zoology Molecular Genetics Additional Questions and Answers

Question 1.
The term‘gene’was coined by ___________
Answer:
Wilhelm Johannsen

Question 2.
Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod and McCarty’s experiment
(c) Hershey-Chase experiment
(d) Urey-Miller’s experiment
Answer:
(c) Hershey-Chase experiment

Question 3.
In Hershey – Chase experiment, the DNA of T 2 phase was made radioactive by using ___________
(a) 32 P
(b) 32 S
(c) 35 P
(d) 32 S
Answer:
(a) 32 P

Question 4.
A nucleoside is composed of ___________
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and Nitrogen base
(d) Sugar, Phosphate and Nitrogenous base
Answer:
(c) Sugar and Nitrogen base

Question 5.
Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer:
(c) Purines have single carbon-nitrogen ring

Question 6.
Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis o f ___________
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer:
(c) Wilkins and Franklin

Question 7.
The term ‘RNA world’ was first used by ___________
Answer:
Walter Gilbert

Question 8.
The distance between two consecutive base pairs in DNA is ___________
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer:
(a) 0.34 nm

Question 9.
If the length of E. coli DNA is 1.36 mm, the number of base pairs is ___________
(a) 0.36 × 10 6 m
(b) 4 × 10 6 m
(c) 0.34 × 10 -9 nm
(d) 4 × 10 -9 m
Answer:
(b) 4 × 10 6 m

Question 10.
Identify the proper sequence in the organisation of eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid – chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer:
(a) Nucleosome – Solenoid – Chromatid

Question 11.
Assertion (A) : Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organisation.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(c) R explains A

Question 12.
Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 13.
Assertion (A) : Semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(a) Both A and R are incorrect

Question 14.
Komberg enzyme is called as _____
Answer:
DNA polymerase I

Question 15.
Replication of DNA occurs at __________ phase of cell cycle.
(a) M
(b) S
(c) G1
(d) G2
Answer:
(b) S

Question 16.
Semi-conservative model of replication was proved by __________
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer:
(c) Meselson and Stahl

Question 17.
How many types of DNA polymerases does an eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer:
(d) Five

Question 18.
Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer:
(c) Unwinding of DNA strand is carried out by topoisomerase

Question 19.
The discontinuously synthesized fragments of lagging strand are called ________
Answer:
Okazaki fragments

Question 20.
Retroviruses possess ________ as genetic material.
Answer:
RNA

Question 21.
Which is NOT a part of transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer:
(b) Operator

Question 22.
Goldberg – Hogness box of eukaryotes is equivalent to ________ of prokaryotes.
Answer:
Pribnow box

Question 23.
Okazaki fragments are joined by the enzyme ________ during DNA replication.
Answer:
DNA ligase

Question 24.

Answer:
(a) A – iv, B – i, C – ii, D – iii

Question 25.
The RNA polymerase of prokaryotes binds with factor to initiate polymerization.
(a) rho
(b) theta
(c) sigma
(d) psi
Answer:
(c) sigma

Question 26.

(a) Capping
(b) Tailing
(c) Splicing
(d) Transcribing
Answer:
(c) Splicing

Question 27.
Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires ‘P’ factor
(d) Split gene feature
Answer:
(d) Split gene feature

Question 28.
Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer:
(d) ii and iv

Question 29.
Capping of mRNA occurs using __________
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer:
(b) Methyl guanosine triphosphate

Question 30.
One of the aspect is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer:
(d) Ambiguous

Question 31.
Which of the triplet codon is not a code of proline?
(i) CCU
(ii) CAU
(iii) CCG
(iv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer:
(b) ii and iv

Question 32.
Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer:
(c) Exons

Question 33.
Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer:
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU

Question 34.
The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) θ
(b) σ
(c) ρ
(d) ∑
Answer:
(c) ρ

Question 35.
In a DNA double strand, if guanine is of 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer:
(b) 20%

Question 36.
Identify the triplet pairs that code for Tyrosine
(a) UUU,UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer:
(b) UAU, UAU

Question 37.

Answer:
A – ii B – i C – iv D – iii

Question 38.
AUG code is for __________
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer:
(d) Methionine

Question 39.
The sequence of bases in coding strand of DNA is G A G T T A G C A G G C, then the sequence of codons in primary transcript is
(a) C U C A U A C G C C C G
(b) C U C A A U C G U C C G
(c) U C A G A U C U G C G C
(d) U U C A A U C G U G C G
Answer:
(b) C U C A A U C G U C C G

Question 40.
The promoter region of eukaryote is __________
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer:
(a) TATAA

Question 41.
Match the following:
(A) AUG – (i) Tyrosine
(B) UGA – (ii) Glycine
(C) UUU – (iii) Methionine
(D) GGG – (iv) Phenylalanine
(a) A – iii B – i C – iv D – ii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – iii D – ii
(d) A – ii B – i C – iv D – iii
Answer:
(d) A – ii B – i C – iv D – iii

Question 42.
__________ number of codons, codes for cystine.
Answer:
Two

Question 43.
In sickle cell anaemia, the __________ codon of β – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Nineth
Answer:
(c) Sixth

Question 44.
Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons donot have tRNA’s
(c) Addition of aminoacid leads to hydrolysis of tRNA
(d) tRNA has four major loops
Answer:
(c) Addition of aminoacid leads to hydrolysis of tRNA

Question 45.
Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer:
(a) Neomycin

Question 47.
The cluster of genes with related function is called _________
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer:
(b) Operon

Question 48.
Repressor protein of Lac operon binds to __________ of operon.
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer:
(b) Operator region

Question 49.
Lac Z gene codes for __________
(a) Permease
(b) transacetylase
(c) β -galactosidase
(d) Aminoacyl transferase
Answer:
(c) β -galactosidase

Question 50.
Lac operon model was proposed by __________
Answer:
Jacob and Monod

Question 51.
Approximate count of base pair in human genome is __________
Answer:
3 × 10 9 bp

Question 52.
Automated DNA sequences are developed by.
Answer:
Frederick Sanger

Question 53.
Which of the chromosome has higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer:
(b) Chromosome 19

Question 54.
Number of genes located in chromosome Y is __________
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer:
(d) 231

Question 55.
How many structural genes are located in lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(b) 3

Question 56.
DNA finger printing technique was developed by
(a) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
Answer:
(b) Alec Jeffreys

Question 57.
In DNA fingerprinting, separation of DNA fragments is done by __________
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer:
(b) Electrophoresis

Question 58.
SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer:
(a) Single nucleotide Polymorphism

Question 59.
Specific sequences of mRNA that are not translated are __________
Answer:
UnTranslated Regions (UTR)

Question 60.
Non-coding or intervening DNA sequence is called __________

Question 61.
_______ Intron is the monomer of DNA.
Answer:
Nucleotide

Question 62.
Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer:
(d) Splicing – Joining of exons with introns

Question 1.
Who proposed One gene – One enzyme hypothesis? Define it.
Answer:
George Beadle and Edward Tatum proposed One gene – One enzyme hypothesis which states that one gene controls the production of one enzyme.

Question 2.
Differentiate nucleoside from nucleotide.
Answer:

  1. Nucleoside: Nucleoside subunit is composed of nitrogenous bases linked to a pentose sugar molecule.
  2. Nucleotide: Nucleotide subunit is composed of nitrogenous bases, a pentose sugar and a phosphate group.

Question 3.
State the key differences between DNA and RNA.
Answer:
DNA:

  1. DNA is made of deoxyribose sugar.
  2. Nitrogenous bases of DNA are Adenine, Guanine, Cytosine and Thymine.
  1. RNA is made of ribose sugar.
  2. Nitrogenous bases of RNA are Adenine, Guanine, Cytosine and Uracil.

Question 4.
Point out the nitrogenous bases of RNA.
Answer:
Adenine, Guanine, Cytosine and Uracil.

Question 5.
What makes the DNA and RNA as acidic molecules?
Answer:
The phosphate functional group (PO4) gives DNA and RNA the property of an acid at physiological pH, hence the name nucleic acid.

Question 6.
Which type of bond is formed

  1. between a purine and pyrimidine base?
  2. between the pentose sugar and adjacent nucleotide?
  1. Purine and pyrimidine bases are linked by hydrogen bonds.
  2. Pentose sugar is linked to adjacent nucleotide by phosphodiester bonds.

Question 7.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

  1. RNA was reactive and hence highly unstable.
  2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
  3. Uracil of RNA is less stable than thymine of DNA.

Question 8.
Name any two viruses whose genetic material is RNA.
Answer:

Question 9.
What are the properties that a molecule must possess to act as genetic material?
Answer:

  1. Self replication
  2. Information storage
  3. Stability
  4. Variation through mutation

Question 10.
How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer:
There are ten base pairs in each turn with a distance of 0.34 x 109m between two adjacent base pairs.

Question 11.
What is a genophore?
Answer:
In prokaryotes such as E. coli though they do not have defined nucleus, the DNA is not scattered throughout the cell. DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

Question 12.
Whqt is nucleosome? How many base pairs are there in a typical nucleosome?
Answer:
The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Question 13.
Expand and define NHC
Answer:

  1. NHC : Non-histone Chromosomal protein.
  2. In eukaryotes, apart from histone proteins, additional set of proteins are required for packing of chromatin at higher level and are referred as non – histone chromosomal proteins.

Question 14.
Differentiate between Heterochromatin and Euchromatin.
Answer:
Heterochromatin:

  1. Region of nucleus where the chromatin are loosely packed and stains light are called Heterochromatin.
  2. Transcriptionally inactive.
  1. Region of nucleus where the chromatin are tightly packed and stains dark are called Euchromatin.
  2. Transcriptionally active.

Question 15.
Which is the widely accepted model of DNA replication? Who has proved it?
Answer:
Semi-conservative replication model. It was proved by Meselson and Stahl in 1958.

Question 16.
Name the chemical substance which is called by the name

  1. DNA polymerase I is also known as Komberg enzyme.
  2. Polynucleotide phosphorylase is also known as Ochoa’s enzyme.

Question 17.
Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer:

  1. DNA Polymerase I Involver in DNA repair mechanism
  2. DNA Polymerase II Involver in DNA repair mechanism
  3. DNA Polymerase III Involver in DNA replication

Question 18.
What is the function of Deoxy nucleotide triphosphate in replication?
Answer:
Deoxy nucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Question 19.
Given below are some events of eukaryotic replication. Name the enzymes involved in the process.

  1. Unwinding of DNA
  2. Joining of Okazaki fragments
  3. Addition of nucleotides to new strand
  4. Correcting the repair

Question 20.
Differentiate leading strand from lagging strand
Answer:
Leading strand:

Question 21.
What are Okazaki fragments?
Answer:
The discontinuously synthesized fragments of the lagging strand are called the Okazaki fragments are joined by the enzyme DNA ligase.

Question 22.
What is a replication fork?
Answer:
At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-Shaped structure called the replication fork. There are two replication forks at each origin.

Question 23.
Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer:
DNA ligase, Topoisomerase (DNA gyrase), Helicase and Nuclease.

Question 24.
Who proposed the central dogma? Write its concept.
Answer:
Francis Crick proposed the Central dogma in molecular biology which states that genetic information flows as follows:

Question 25.
Define transcription and name the enzyme involved in this process.
Answer:
The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA dependent RNA polymerase.

Question 26.
What is TATA box? State its function.
Answer:
In eukaryotes, the promoter has AT rich regions called TATA box or Goldberg-Hogness box. It acts as a binding site for RNA polymerase.

Question 27.
Structural gene of eukaryotes differ from prokaryotes. How?
Answer:
In eukaryotes, the structural gene is monocistronic coding for only one protein whereas in prokaryotes the structural gene is polycistronic coding for many proteins.

Question 28.
What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer:
Bacterial (prokaryotic) RNA polymerase consists of two major components, the core enzyme and the sigma subunit. The core enzyme (β1, β, and α) is responsible for RNA synthesis ” whereas a sigma subunit is responsible for recognition of the promoter.

Question 29.
Distinguish between exons and introns.
Answer:

  1. Exons: Expressed sequences (Coding sequences) of an eukaryotic gene
  2. Introns: Interveining sequences (non-coding sequences) of an eukaryotic gene

Question 30.
Define splicing.
Answer:
The process of removing introns from hnRNA is called splicing.

Question 31.
What is capping and tailing?
Answer:
In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5’ end of hnRNA, whereas adenylate residues (200-300) (Poly A) are added at the 3’ end in tailing.

Question 32.
If a double stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer:
Cytdsine = 20, hence Guanine = 20
As per ChargafFs rule (A+T) = (G+C) =100
Percent of Thymine + Adenine = 20 + 20 = 100
(T + A) = (20 + 20) =100
(T + A)=100-(20 + 20)
T +A = 100 – 40
T + A = 60
Therefore the percent of Adenine will be 60/2 = 30%.

Question 33.
Mention the dual functions of AUG.
Answer:
AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

Question 34.
How many codons are involved in termination of translation. Name them.
Answer:
Three codons terminate translation process. They are UAA, UAG and UGA.

Question 35.
Degeneracy of codon – comment.
Answer:
A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

Question 36.
Point out the exceptional categories to universality of genetic code.
Answer:
Exceptions to universal nature of genetic code is noticed in prokaryotic mitochondrial and chloroplast genomes.

Question 37.
What are non-sense codons?
Answer:
UGA, UAA and UAG are the non-sense codons, which terminates translation.

Question 38.
Name the triplet codons that code for

Question 39.
Why hnRNA has to undergo splicing?
Answer:
Since hnRNA contains both coding sequences (exons) and non-coding sequences (introns) it has to undergo splicing to remove introns.

Question 40.
State the role of following codons in translation process

Question 41.
Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
Answer:
3’AUGAAAGAUGGGUAA5’
Methionine – Lysine – Aspartic acid – Glycine

Question 42.
Name the four codons that codes valine.
Answer:
GUU, GUC, GUA and GUG.

Question 43.
The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer:
The complementary strand has ATCGTACTA.

Question 44.
Why t-RNA is called as adapter molecule?
Answer:
The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule.

Question 45.
What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer:
The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl- tRNA (charged tRNA). Aminoacylation is catalyzed by an enzyme aminoacyl – tRNA synthetase.

Question 46.
What are UTR’s?
Answer:
mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5’ end (before start codon) and at 3’ end (after stop codon).

Question 47.
What is S – D sequence?
Answer:
The 5’ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgamo sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 48.
Define translation unit.
Answer:
A translation unit in mRNA is the sequence of RNA that is flanked by the start codon on 5’ end and stop codon on 3’ end and codes of polypeptide.

Question 49.
Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer:
Tetracycline inhibits binding between aminoacyl tRNA and mRNA.Streptomycin inhibits initiation of translation and causes misreading.

Question 50.
At what stage, does the gene expression is regulated?
Answer:
Gene expression can be controlled or regulated at transcriptional or translational levels.

Question 51.
What is a operon? Give example.
Answer:
The cluster of genes with related functions is called operon.
E.g: lac operon in E.coli.

Question 52.
Considering the lac operon of E.coli, name the products of the following genes.

  1. i gene – repressor protein
  2. lac Z gene – fS-galactosidase
  3. Lac Y gene – Permease
  4. lac a gene – transacetylase

Question 54.
Name the human chromosome that has

  1. Chromosome 1 has maximum number of genes (2968 genes)
  2. Chromosome Y has least genes (231 genes)

Question 55.
What are SNPs? Mention its uses.
Answer:
SNPs : Single nucleotide polymorphism. It helps to find chromosomal locations for disease associated sequences and tracing human history.

Question 56.
Mention any four areas where DNA fingerprinting can be used.
Answer:

  1. Forensic analysis
  2. Pedigree analysis
  3. Conservation of wild life
  4. Anthropological studies

Question 57.
Classify nucleic acid based on sugar molecules.
Answer:
There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Question 58.
Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer:
Both purines and pyrimidines are nitrogen bases. The purine bases Adenine and Guanine have double carbon – nitrogen ring, whereas cytosine and thymine bases have single carbon nitrogen ring.

Question 59.
How 5’ of DNA differ from its 3’?
Answer:
The 5’ of DNA refers to the carbon in the sugar to which phosphate (P04V) functional group is attached. The 3’ of DNA refers to the carbon in the sugar to which a hydroxyl (OH) group is attached.

Question 60.
State Chargaff’s rule.
Answer:
According to Erwin Chargaff,

  1. Adenine pairs with Thymine with two hydrogen bonds.
  2. Guanine pairs with Cytosine with three hydrogen bonds.

Question 61.
Chemically DNA is more stable than RNA – Justify.
Answer:
In DNA, the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2 OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Question 62.
Write in simple about semi-conservative mode of DNA replication.
Answer:
Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesised and complementary to the parent strand.

Question 63.
Draw a simplified diagram of nucleosome and label it.
Answer:

Question 64.
What is a primer?
Answer:
A primer is a short stretch of RNA. It initiates the formation of new strand. The primer produces 3’-OH end on the sequence of ribonucleotides, to which deoxyribonucleotides are added to form a new strand.

Question 65.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.

  1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.
  2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 66.
What do you mean by a template strand and coding strand?
Answer:
DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→ 5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand.

Question 67.
Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer:

  1. Sigma factor is responsible for initiation of transcription.
  2. Rho factor is responsible for termination of transcription.

Question 68.
Name the major RNA types of prokaryotes and mention their role.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role
during translation.

Question 69.
Define genetic code.
Answer:
The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism. This specific order of base pairs is called genetic code.

Question 70.
Explain Wobble hypothesis.
Answer:
Wobble Hypothesis is proposed by Crick (1966) which states that tRNA anticodon has the ability to wobble at its 5’ end by pairing with even non-complementary base of mRNA codon.’ According to this hypothesis, in codon-anticodon pairing the third base may not be complementary.

The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy.

Question 71.
Explain the nature of eukaryotic ribosome.
Answer:
The ribosomes of eukaryotes (80 S) are larger, consisting of 60 S and 40 S sub units. ‘S’ denotes the sedimentation efficient which is expressed as Svedberg unit (S). The larger subunit in eukaryotes consist of a 23 S RNA and 5Sr RNA molecule and 31 ribosomal proteins. The smaller eukaryotic subunit consist of 18Sr RNA component and about 33 proteins.

Question 72.
Expand and define ORF.
Answer:
Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF).

Question 73.
What are the components of initiation complex of prokaryotic translation?
Answer:
Initiation of translation in E. coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (f met -1 RNA f met ), three proteinaceous initiation factors (IF 1, IF2, IF3), GTP (Guaniner Tri Phosphate) and Mg 2+ .

Question 74.
Explain the components of operon.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists
of one or more structural genes and an adjacent operator gene that controls transcriptional
activity of the structural gene.

  1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
  2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase I binds to the promoter prior to the initiation of transcription.
  3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 75.
Describe Hershey and Chase experiment. What is concluded by their experiment?
Answer:
Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T2 is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine). Hershey and Chase designed an experiment using radioactive isotopes of Sulphur ( 35 S) and phosphorus ( 32 P) to keep separate track of the viral protein and nucleic acids during the infection process.

The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes 35 S or 32 P. The bacteriophage that grew in the presence of 35 S had labelled proteins and bacteriophages grown in the presence of 32 P had labelled DNA. The differential labelling thus enabled them to identity DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material. Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phase particles.

It was observed that only 32 P was found associated with bacterial cells and 35 S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only 32 P and not 35 S. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

Question 76.
Explain the properties of DNA that makes it an ideal genetic material.
Answer:
1. Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

2. Stability: It should he stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary.

if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

3. Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

4. Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 77.
How the DNA is packed in an eukaryotic cell? ft
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (HI) that is exposed to enzymes.

The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule. Chromatin lacking HI has a beads-on-a-string appearance in which DNA inters and leaves the nucleosomes at random places. HI of one nucleosome can interact with 33l of the neighbouring nucleosomes resulting in the further folding of the fibre.

The chrof&atin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucfeosbme, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different HI molecules. DNA is a solenoid and packed about,%)_folds. The hierarchical nature of chromosome structure is illustrated.

Additional set of pteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In*,a typical nucleus, some regions of chromatin are Ibosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is,-tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

Question 78.
Meselson and Stahl’s experiment proved the semi-coflBptervation mode of DNA replication. Explain.
Answer:
The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi-conservative, conservative and dispersive replications. In their experiment, they grew two cultures of E.coli for many generations in separate media. The ‘heavy’ culture was grown in a medium in which the nitrogen source (NH4CI) contained the heavy isotope 15 N and the ‘ light’ culture was grown in a medium in which the nitrogen source contained light isotope 14 H for many generations. At the end of growth, they observed that the bacterial DNA in the heavy culture contained only 15 N and in the light culture only 14 N. The heavy DNA could be distinguished from light DNA ( 15 N from 14 N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in thtytwo cultures settled into two distinct and separate bands (hybrid DNA).

The heavy culture ( 15 N) was then transferred into a medium that had only NH4CI, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples,and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick’s semi – conservative replication hypothesis.

Question 79.
Give a detailed account of a transcription unit.
Answer:
A transcriptional unit in DNA is defined by three regions, a promoter, the structural gene and a terminator. The promoter is located towards the 5 ’ end. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3’ end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing. In eukaryotes the promoter has AT rich regions called TATA box (Goldberg- Hogness box) ‘ and in prokaryotes this region is called Pribnow box.

Besides promoter, eukaryotes also require an enhancer. The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each mRNA carries only a single gene and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Question 80.
Explain the transcription process in prokaryotes with needed diagram.
Answer:
In prokaryotes, there are three major types of RNAs:
mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation. There is a single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation).

The polymerases binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerases in a template depended fashion following the rule of complementarity. After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the, nascent RNA falls off, so also the RNA polymerase. The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (a) and termination factor rho (p) to initiate and terminate the transcription, respectively.

Association of RNA with these factors instructs the RNA polymerase either to initiate or terminate the process of transcription. In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment sincethere is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not separated from other cell organelles by a nuclear membrane consequently transcription and translation can be coupled in bacteria.

Question 81.
Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

  1. The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).
  2. The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.
  3. A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUTJ and GUC represents only two codons.
  4. It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes.
  5. A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.
  6. Non-ambiguous code means that one codon will code for one amino acid.
  7. The code is always read in a fixed direction i.e. from 5’→3’ direction called polarity.
  8. AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.
  9. UAA, UAG (tyrosine) and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 82.
Mutations on genetic code affects the phenotype. Describe with example.
Answer:
The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens. A well studied example is sickle cell anaemia in humans which results from a point mutation of an allele of β-haemoglobin gene (βHb).

A haemoglobin molecule consists of four polypeptide chains of two types, two a chains and two P-chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The jruman blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in haemoglobin is due to a single base substitution at the sixth codon of the beta globingene from GAG to GTG in p -chain of haemoglobin.

It results in a change of amino acid glufeniic acid to valine at the 6th position of the p -chain. This is the classical example of point mutation that results in the change of amino acids residue glutamic acid to valine. The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.

Question 83.
Explain the mechanism of AteArperon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes – permease, P-galactosidase (P-gat) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, Pjgglactosidase brings about hydrolysis of lactose to glucose and galactose, while transacety gtransfers acetyl group from acetyl Co A to P-galactosidase. The lac operon consists of one-regulator gene (T gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for P-gaiaqtttsidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli. In lac a polycistronic structural gene is regulated by a common promoter and regulatory genfc When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, P-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it.

The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism. This regulation of lac operon by the repressor is an example of negative control of transcription initiation.

Question 84.
What are the objectives of Human Genome project?
Answer:
The main goals of Human Genome Project are as follows:

  1. Identify all the genes (approximately 30000) in human DNA.
  2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
  3. To store this information in databases.
  4. Improve tools for data analysis.
  5. Transfer related technologies to other sectors, such as industries.
  6. Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 85.
Write the salient features of Human Genome Project.
Answer:

  1. Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
  2. An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
  3. The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
  4. Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
  5. The chromosomal organization of human genes shows diversity.
  6. There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
  7. Functions for over 50 percent of the discovered genes are unknown.
  8. Less than 2 percent of the genome codes for proteins.
  9. Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
  10. Chromosome 1 has 2968 genes, whereas chromosome ’Y’ has 231 genes.
  11. Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotidepolymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

Question 86.
Describe the principle involved in DNA fingerprinting technique.
Answer:
The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and finger prints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences. In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called as variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers.

DNA finger printing involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A: T rich or G : C rich), length of segment and number of repetitive units, the satellite DNA is classified into many sub – categories such as micro-satellites and mini satellites, etc.

These sequences do not code for any proteins, but they form a large portion of human genome. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting. DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Question 87.
Draw a flow chart depicting the steps of DNA finger printings technique
Answer:

Higher Order Thinking Skills (HOTs) Questions

Question 1.
A mRNA strand has a series of triplet codons of which the first three codons are given below
(a) AUG
(b) UUU
(c) UGC
(i) Name the amino acid encoded by these triplet codons.
(ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) AUG codes for Methionine
UUU codes for Phenylalanine
UGC codes for Cysteine
(ii) TAC sequence of DNA is transcribed to AUG
AAA sequence of DNA is transcribed to UUU
ACG sequence of DNA is transcribed to UGC

Question 2.
Given below are the structures of tRNA molecules which are involved in translation process. In one tRNA, codon is mentioned but not the amino acid. In another tRNA molecule, amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:

Question 3.
A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total number of nucleotides does that DNA fragment contains? Explain.
Answer:
128 nucleotides. Adenine always pair Thymine base. If there are 32 adenine bases then there must be 32 Thymine bases. Similarly cytosine pairs with guanine. If cytosine bases are 32 in number the guanine bases will be equal to cytosine. So it make a total of 128 nucleotides.

Question 4.
Following is a DNA sequence representing a part of gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A.


ORF Finder:

The ORF finder is a program available at NCBI website. It identifies the all open reading frames or the possible protein coding region in sequence. It shows 6 horizontal bars corresponding to one of the possible reading frame. In each direction of the DNA there would be 3 possible reading frames. So total 6 possible reading frame (6 horizontal bars) would be there for every DNA sequence. The 6 possible reading frames are +1, +2, +3 and -1, -2 and -3 in the reverse strand. The resultant amino acids can be saved and search against various protein databases using blast for finding similar sequences or amino acids. The result displays the possible protein sequence and the length of the open reading frame etc.


Key Discovery About How Genes Turn On and off – “Critical Implications for Human Health”

Human bodies have roughly 30,000 genes dictating not only how we look, but also critical biological processes. Now, a Florida State University and Australia National University research team has discovered a key aspect of gene regulation and ultimately how that process is implicated in cancer.

Jonathan Dennis, an associate professor of biological science at FSU, and David Tremethick, a professor at Australia National University, have published a new paper in Nature Communications that reveals key information about a gene’s controlling region – where proteins attach to turn genes on or off. The researchers found that the way this region is packaged dictates how genes are either expressed or restricted.

The packaging refers to all of the characteristics of how and where these proteins attach. That process is critical to human biology, Dennis noted.

“When the wrong thing binds, you get inappropriate physiology, in some cases, cancer,” he said.

The new information challenges the current models for how a gene is expressed by revealing that there are many different ways a promoter can be packaged to either allow or restrict the expression of a gene.

A protein called H2A.Z plays an important role in regulating this packaging of genes in different ways. The researchers found that one important role of H2A.Z in gene regulation is to ensure that only the proper regulatory factors have access to gene promoters.

“H2A.Z is a type of protein called a histone variant,” said Lauren Cole, a former FSU doctoral student and the first author on the paper. “Because histone variants play an important role in gene regulation, this work leads to an expanded understanding of the human genome.”

Tremethick said the finding underscores how much work is left to be done to understand the human genome and how this finding can advance field forward.

“Although it has been nearly 20 years since the human genome was sequenced, how this genomic information is selectively utilized to direct patterns of gene expression underpinning cell fate decisions still remains poorly understood,” Tremethick said. “While there is still much work to be done, our study will help move the field forward to get a better understanding of how our genes are expressed at the right time and place, which has critical implications for human health.”

Reference: “Multiple roles of H2A.Z in regulating promoter chromatin architecture in human cells” by Lauren Cole, Sebastian Kurscheid, Maxim Nekrasov, Renae Domaschenz, Daniel L. Vera, Jonathan H. Dennis and David J. Tremethick, 5 May 2021, Nature Communications.
DOI: 10.1038/s41467-021-22688-x

Other authors on the paper are Sebastian Kurscheid, Maxim Nekrasov and Renae Domaschen of Australian National University, and Daniel Vera, a former FSU graduate student who is now a research fellow at Harvard Medical School.

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2 Comments on "Key Discovery About How Genes Turn On and off – “Critical Implications for Human Health”"

Babu G. Ranganathan*
(B.A. Bible/Biology)

HOW DOES DNA TURN A CELL INTO A SHEEP, OR A BIRD, OR A HUMAN?

When you divide a cake, the cake never gets bigger. However, when we were just a single cell and that cell kept dividing we got bigger. New material had to come from somewhere. That new material came from food.

Just as the sequence of various letters and words in human language communicate a message and direct workers to build and assemble something so, too, the sequence of various molecules in our DNA (our genes or genetic code) directed the molecules from our mother’s food, that we received in the womb, to become new cells, eventually forming all the tissues and organs of our body.

When you feed a cat your food the cat’s DNA will direct the food molecules to become the cells, tissues, and organs of a cat, but your DNA will turn the same food into human cells, tissues, and organs.

What we call “genes” are actually segments of the DNA molecule. When you understand how your DNA works, you’ll also understand how egg yolks can turn into chickens. Read my popular Internet article: HOW DID MY DNA MAKE ME? Just google the title to access the article.

This article will give you a good understanding of how DNA, as well as cloning and genetic engineering. You also learn that so-called “Junk DNA” isn’t junk at all. You will learn why it is not rational to believe that DNA code could have arisen by chance. Science points (not proves, but points) to an intelligent cause for DNA code.

What about genetic and biological similarities between species? Genetic information, like other forms of information, cannot happen by chance, so it is more logical to believe that genetic and biological similarities between all forms of life are due to a common Designer who designed similar functions for similar purposes. It doesn’t mean all forms of life are biologically related! Only genetic similarities within a natural species proves relationship because it’s only within a natural species that members can interbreed and reproduce.

Nature cannot build DNA code from scratch. It requires already existing DNA code to direct and bring about more DNA code or a genetic engineer in the laboratory using intelligent design and highly sophisticated technology to bring DNA code into existence from scratch. Furthermore, RNA/DNA and proteins are mutually dependent (one cannot come into existence without the other two) and cannot “survive” or function outside of a complete and living cell. DNA code owes its existence to the first Genetic Engineer – God!

Protein molecules require that various amino acids come together in a precise sequence, just like the letters in a sentence. If they’re not in the right sequence the protein won’t function. DNA and RNA require for various their various nucleic acids to be in the right sequence.

Furthermore, there are left-handed and right-handed amino acids and there are left-handed and right-handed nucleic acids. Protein molecules require for all their amino acids to be left-handed only and in the right sequence. DNA and RNA require for all their nucleic acids to be right-handed and in the right sequence. It would take a miracle for DNA, RNA, and proteins to arise by chance!

Mathematicians have said any event in the universe with odds of 10 to 50th power or greater is impossible! The probability of just an average size protein molecule (with its amino acids in the right sequence) arising by chance is 10 to the 65th power. Even the simplest cell is made up of many millions of various protein molecules along with and DNA/RNA..

The late great British scientist Sir Frederick Hoyle calculated that the odds of even the simplest cell coming into existence by chance is 10 to the 40,000th power! How large is this? Consider that the total number of atoms in our universe is 10 to the 82nd power.

Also, so-called “Junk DNA” isn’t junk. Although these “non-coding” segments of DNA don’t code for proteins, they have recently been found to be vital in regulating gene expression (i.e. when, where, and how genes are expressed, so they’re not “junk”). Also, there is evidence that, in certain situations, they can code for proteins through the cell’s use of a complex “read-through” mechanism.

Visit my latest Internet site: THE SCIENCE SUPPORTING CREATION (This site answers many arguments, both old and new, that have been used by evolutionists to support their theory)

Author of the popular Internet article, TRADITIONAL DOCTRINE OF HELL EVOLVED FROM GREEK ROOTS

*I have given successful lectures (with question and answer period afterwards) defending creation before evolutionist science faculty and students at various colleges and universities. I’ve been privileged to be recognized in the 24th edition of Marquis “Who’s Who in The East.”


DNA Code for Insulin

Below are two partial sequences of DNA bases (shown for only one strand of DNA) Sequence 1 is from a human and sequence 2 is from a cow. In both humans and cows, this sequence is part of a set of instructions for controlling the production of a protein. In this case, the sequence contains the gene to make the protein insulin. Insulin is necessary for the uptake of sugar from the blood. Without insulin, a person cannot use digest sugars the same way others can, and they have a disease called diabetes.

  1. Using the DNA sequence given in table 1, make a complimentary RNA strand for the human. Write the RNA directly below the DNA strand (remember to substitute U’s for T’s in RNA).
  2. Repeat step 1 for the cow. Write the RNA directly below the DNA strand in table 2.
  3. Use the codon table in your book to determine what amino acids are assembled to make the insulin protein in both the cow and the human. Write your amino acid chain directly below the RNA sequence.

Sequence 1 ­ Human
DNA C C A T A G C A C G T T A C A A C G T G A A G G T A A
RNA
Amino Acids

Sequence 1 ­ Cow
DNA C C G T A G C A T G T T A C A A C G C G A A G G C A C
RNA
Amino Acids

1. The DNA sequence is different for the cow and the human, but the amino acid chain produced by the sequence is almost the same. How can this happen?

2. Diabetes is a disease characterized by the inability to break down sugars. Often a person with diabetes has a defective DNA sequence that codes for the making of the insulin protein. Suppose a person has a mutation in their DNA, and the first triplet for the gene coding for insulin is C C C (instead of C C A). Determine what amino acid the new DNA triplet codes for. Will this person be diabetic?

3. What if the first triplet was C A A ?

4. How is it that a code consisting of only four letters, as in DNA ( A, T, G, C ) can specify all the different parts of an organism and account for all the diversity of organisms on this planet?

DNA sequences are often used to determine relationships between organisms. DNA sequences that code for a particular gene can vary widely. Organisms that are closely related will have sequences that are similar. Below is a list of sequences for a few organisms:

Human CCA TAG CAC CTA
Pig CCA TGG AAA CGA
Chimpanzee CCA TAA CAC CTA
Cricket CCT AAA GGG ACG

5. Based on the sequences, which two organisms are most closely related?

6. An unknown organism is found in the forest, and the gene is sequenced, and found to be C C A T G G A A T C G A , what kind of animal do you think this is?


Bond booster

Scientists in the Soviet Union were the first to discover Z-DNA, in the late 1970s, in a phage called S-2L, which infects photosynthetic bacteria 4 . They found that the phage DNA behaved oddly when its two helical strands were melted apart. The bond that forms between G and C bases breaks at a higher temperature, compared with that joining A and T, and the phage’s DNA behaved as if it was made primarily from G and C. But further analysis by the Soviet team showed that the phage had replaced A with Z, which formed a stronger bond with T.

“It looked like something transgressive,” says Philippe Marlière, an inventor and geneticist at the University of Evry, France, who led one of the Science studies. “Why did this phage have a special base like this?”

Scientists glimpse oddball microbe that could help explain rise of complex life

Follow-up studies showed that S-2L’s heartier genome was resistant to DNA-chomping enzymes and other anti-phage defences that bacteria wield. But researchers didn’t know how the Z-DNA system worked or whether it was common. Z-DNA is only one of a host of modifications known to exist in phage DNA.

To answer those questions, a team led by Marlière and Pierre-Alexandre Kaminski, a biochemist at the Pasteur Institute in Paris, sequenced the phage’s genome in the early 2000s. They found a gene that’s potentially involved in one step of making Z-DNA, but not in others. But the sequence had no matches in genomic databases at the time, and the team’s quest to understand the basis for Z-DNA hit a dead end.

Marlière and his colleagues patented the S-2L genome, but also made it public, and he continued to scour genomic databases. Finally, in 2015, the team got a hit: a phage that infects aquatic bacteria of the genus Vibrio harboured a gene that matched a stretch of S-2L’s genome. The gene encoded an enzyme that resembled one that bacteria use to make adenine. “It was an exhilarating moment,” says Marlière.

In 2019, Zhao’s team found similar database matches. Both teams showed that the phages all had a gene named PurZ. This codes for an enzyme that plays an early but crucial part in making the Z nucleotide from a precursor molecule that is present in bacterial cells. They then identified additional enzymes — encoded in the genomes of bacteria that the phages infect — that complete the pathway.

DNA's secret weapon against knots and tangles

But a key question lingered. The enzymes that the teams identified produced the raw ingredient for Z-DNA — a molecule called dZTP — but that didn’t explain how phages insert the molecule into DNA strands, while excluding A bases (in the form of a chemical called dATP).

Here, the teams’ conclusions differed slightly. Alongside PurZ in the Vibrio phage’s genome sits a gene that makes an enzyme called a polymerase, which copies DNA strands. Marlière and Kaminski found that the phage polymerase incorporates dZTP into DNA, while cutting out any A bases that were introduced. “This explained to us why A was excluded,” says Kaminski. “This was really spectacular.”

Zhao thinks this isn’t the whole story. Her work suggests that another phage enzyme is needed, one that breaks up dATP but preserves dZTP inside cells. Her team found that increasing dZTP levels relative to those of dATP was enough to trick a cell’s own polymerase into making Z-DNA.


Difference Between Protein Synthesis and DNA Replication

Proteins and DNA provide the most fundamental layout to maintain the life on Earth. In fact, proteins determine the shape and functions of the organisms while DNA keeps the information needed for that. Hence, synthesis of protein and DNA replication could be understood as extremely important processes that take place in the living cells. Both these processes start from the nucleotide sequence of the nucleic acid strand, but those are different pathways. The important steps of both processes are explained, and the differences between them are discussed in this article.

Protein Synthesis

Protein synthesis is a biological process that takes place inside the cells of organisms in three main steps known as Transcription, RNA processing, and Translation. In the transcription step, nucleotide sequence of the gene in the DNA strand is transcribed into RNA. This first step is highly similar to the DNA replication except the result is a strand on RNA in protein synthesis. The DNA strand being dismantled with DNA helicase enzyme, RNA polymerase is attached at the specific place of the start of the gene known as promoter, and RNA strand is synthesized along the gene. This newly formed RNA strand is known as the messenger RNA (mRNA).

The mRNA strand takes the nucleotide sequence to the ribosomes for the RNA processing. Specific tRNA (transfer RNA) molecules will recognize the relevant amino acids in the cytoplasm. After that, tRNA molecules are attached to the specific amino acids. In each tRNA molecule, there is a sequence of three nucleotides. A ribosome in the cytoplasm is attached to the mRNA strand, and the starting codon (the promoter) is identified. The tRNA molecules with the corresponding nucleotides for the mRNA sequence are moved into the large subunit of the ribosome. As the tRNA molecules come to the ribosome, the corresponding amino acid is bonded with the next amino acid in the sequence through a peptide bond. This last step is known as translation indeed, this is where the actual protein synthesis takes place.

The shape of the protein is determined through the different types of amino acids in the chain, which were attached to tRNA molecules, but tRNA are specific to the mRNA sequence. Hence, it is clear that the protein molecules depict the information stored in the DNA molecule. However, protein synthesis could be initiated from an RNA strand, as well.

DNA Replication

DNA replication is the process of producing two identical DNA strands from one, and it involves a series of processes. All these processes take place during the S phase of the Interphase of cell cycle or cell division. It is an energy consuming process and primarily three main enzymes known as DNA helicase, DNA polymerase, and DNA ligase are involved in regulating this process. First, DNA helicase dismantles the double helix structure of the DNA strand by breaking the hydrogen bonds between the nitrogenous bases of the opposing strands. This dismantling starts from an end of the DNA strand and not from the middle. Therefore, DNA helicase could be regarded as a restriction exonuclease.

After exposing the nitrogenous bases of the single stranded DNA, the corresponding Deoxyribonucleotides are arranged according to the base sequence and the respective hydrogen bonds are formed by DNA polymerase enzyme. This particular process takes place on both DNA strands. Finally, the phosphodiester bonds are formed between successive nucleotides, to complete the DNA strand using DNA ligase enzyme. At the end of all these steps, two identical DNA strands are formed from only one mother DNA strand.

Difference between Protein Synthesis and DNA Replication


DISCUSSION

Monomeric PLD-superfamily enzymes like human Tdp1 and PLD from Streptomyces species ( 12, 13) are bi-lobed monomers ( Figure 1A) which contain in the active site two His residues from duplicated ‘HXK’ sequence motifs located distantly in the protein chain. The active-site histidines of these enzymes are not equivalent and perform pre-defined roles in catalysis. One particular His mounts the nucleophilic attack on the scissile phosphate to make a covalent intermediate ( Figure 1B). Not surprisingly, its replacement by site-directed mutagenesis renders the enzyme completely inactive ( 13, 16, 26). The other His plays a supporting role—it protonates the leaving group during the formation the covalent intermediate and subsequently facilitates the hydrolysis of the phosphohistidine linkage. Mutations of the latter histidine residue often compromise catalytic activity and result in the accumulation of the covalent intermediate ( 27).

The structurally characterized PLD-family nucleases—Nuc and the restriction endonuclease BfiI ( 10, 14)—are homodimers which contain a single active site structurally similar to that of human Tdp1 and PLD from Streptomyces species. However, in contrast to monomeric PLD enzymes, the active site of BfiI is fully symmetric, as it contains two His residues related by the 2-fold symmetry axis of the dimer, each donated by one enzyme subunit ( Figure 1A). This precludes the assignment of the individual roles of each His residue in catalysis. To solve this problem, we disrupted the 2-fold symmetry intrinsic to BfiI by constructing heterodimeric forms of the enzyme ( Figure 2).

The roles of active-site histidines in catalysis

WT BfiI forms the covalent intermediate on truncated phosphodiester and 3′-phosphorothiolate substrates with comparable rates (2.1 and 7.7 s −1 , respectively: Figure 3E), despite the substantially more acidic leaving group of the 3′-phosphorothiolate substrate [the pKa values of the 3′-SH and 3′-OH groups are ∼11 and ∼16, respectively ( 28)], suggesting that protonation of the 3′-leaving group is not a rate-determining factor for WT BfiI. The ability of BfiI to cleave the 3′-phosphorothiolate linkage more rapidly than the all-oxygen substrate sharply contrasts with most metal-dependent nucleases ( 5, 29, 30). These enzymes, unlike the metal independent BfiI, are inhibited by the 3′-S substitution due to its impaired interaction of Mg 2+ ions with sulfur however, some of these enzymes are rescued by the more thiophilic Mn 2+ ion ( 5, 29).

The replacement of one of the active-site histidines with alanine, in the WT/H105A heterodimer, resulted in a dramatic 10 6 -fold decrease in the rate of cleavage of the oxyester bond in the 14/15 oligoduplex ( Figure 3E). The residual activity could have been due to a small amount of WT BfiI present in the sample of heterodimer but different preparations of the heterodimer, including the alternative variants WT(6His)/H105A and H105A(6His)/WT, all gave the same low level of activity. Therefore, the observed activity is most likely intrinsic to the WT/H105A heterodimer. Thus, the second His105 residue in the active site of BfiI accelerates the formation of the covalent intermediate on the phosphodiester substrate by a factor of at least 10 6 .

Strikingly, the heterodimer cleaved the 3′-phosphorothiolate linkage in the 14/15s duplex 10 5 -fold more rapidly than the phosphodiester group in the original 14/15 substrate ( Figure 3E). The resultant rate of covalent intermediate formation by the heterodimer (0.19 s −1 ) is only 40-fold lower than that for the WT enzyme on the same substrate (7.7 s −1 , Figure 3E). Furthermore, the rate enhancement over the all-oxygen substrate by a factor of 10 5 coincides with the pKa difference between the 3′-OH and 3′-SH leaving groups [∼5 units ( 28)]. These observations argue that one of the two H105 residues in the WT homodimer of BfiI protonates the 3′-leaving group during the first reaction step ( Figure 1B): upon its removal, in the WT/H105A heterodimer, the stability of the conjugate base of the 3′-leaving group becomes a major factor governing the reaction rate. The difference in rate between the heterodimeric and WT enzymes on the 3′-phosphorothiolate substrate implies that even the relatively acidic 3′-thio leaving group must be protonated to achieve the maximum rate of covalent intermediate formation.

The second step in the reaction, the hydrolysis of the covalent enzyme–DNA intermediate ( Figure 1B), is an extremely rapid process for WT BfiI (k2 = 170 s −1 , Figure 3E). However, the rate is reduced by a factor of 17 000 for the WT/H105A heterodimer (k2 = 0.010 s −1 ). The dramatic effect of the H105A substitution confirms the direct involvement of the second active-site histidine in the hydrolysis of the covalent phosphohistidine intermediate presumably, the second histidine activates the water molecule that hydrolyzes the intermediate by removing a proton ( Figure 1B).

The single H105A substitution also reverses the ratio of the reaction rates for the formation (k1) and decay (k2) of the covalent intermediate. For WT BfiI on the 14/15s substrate, the ratio of k1/k2 is 0.05, i.e. the covalent intermediate is formed 20 times more slowly than it is hydrolyzed. For the WT(6His)/H105A and H105A/WT-N heterodimers, the ratio of k1/k2 is ≥20, which leads to the accumulation of the covalent intermediate during the reaction ( Figures 3C and 4A). Biochemical analysis of the low-mobility species postulated to be the covalent intermediate ( Supplementary Data ) indicated that this was the expected phosphohistidine adduct. The covalent intermediate is formed only by the wild type but not the H105A subunit of the heterodimer ( Supplementary Data ). Moreover, as expected for a phosphohistidine compound ( 22, 24), the BfiI–DNA adduct is stable at alkaline pH but decomposes in acid ( Supplementary Data ).

A novel mechanism for double-stranded DNA cleavage

Nucleases that cut double-stranded DNA often contain two identical subunits related by rotational symmetry, so that the active site from one subunit cleaves the 5′–3′ strand while that from the oppositely-oriented subunit attacks the anti-parallel 3′-5′ strand ( 7). However, this strategy cannot be generalized for all nucleases that act on double-stranded DNA. A number of enzymes including the homing endonuclease I-TevI ( 31), the RecBCD complex of E. coli ( 32) and the BfiI restriction enzyme ( 17) all utilize single active site to cut both DNA strands, despite their opposite polarities.

The intron-encoded endonuclease I-TevI is a monomer and contains a single active site but it cuts both DNA strands at the recipient site for intron homing, leaving in both cases products with 3′-hydroxyl and 5′-phosphate termini ( 33). It is thought that it first cleaves its target phosphodiester bond in the bottom strand and then distorts the DNA to guide into the active site the scissile phosphate from the top strand ( 31). However, it is not yet clear how its single active site can accommodate and cut phosphodiester bonds from both the 3′–5′ and the 5′–3′ strands of the DNA, as in both cases it has to displace the leaving group on the 3′ side of the phosphorous at the scissile bond: i.e., in opposite directions on the 3′–5′ compared to the 5′–3′ strand. Moreover, given the crystal structure of the catalytic domain of I-TevI, alternative reaction schemes, including transient dimerization, cannot be excluded ( 34). The mode of action of another monomeric endonuclease, FokI, involves transient dimerization ( 35, 36), to give a protein assembly at the recognition site with two catalytic domains juxtaposed in anti-parallel alignment ( 37, 38), which each cut one strand of the DNA. In this case, the 1° monomer bound directly to the recognition site cleaves the bottom strand while the 2° monomer recruited to the site by protein–protein interactions cuts the top strand ( 25), but the symmetry within the dimer of catalytic domains ( 37) enables one to cut the 3′–5′ strand and the other the 5′–3′ strand.

The E. coli RecBCD enzyme acts in the repair of double-stranded DNA breaks as a trimeric protein with multiple catalytic activities that include two helicase functions, both 3′→5′ and 5′→3′ in the B and D subunits respectively and a single endonuclease function, located in B from where it degrades both strands ( 32). To account for how the nuclease cleaves both of the newly unwound strands despite their opposite polarities, it was suggested that the nascent 3′-terminus generated by RecB progresses directly into the nuclease centre, which is also in B, while the nascent 5′-terminus generated by RecD forms a loop before entering the nuclease centre with the same 3′–5′ orientation as the 3′-strand ( 39). However, this model has yet to be confirmed experimentally, though it can readily be reconciled to the crystal structure of RecBCD ( 40).

The BfiI restriction enzyme employs yet another strategy. It had been shown previously that it uses a single active site to cut both DNA strands downstream of its recognition site in sequential steps, in a fixed order first the bottom and only then the top strand ( 17). The BfiI endonuclease contains two symmetrically-positioned His residues at the active site, so it was proposed that BfiI cuts one strand by using the histidine from one subunit as the nucleophile and that from the other subunit as the proton donor/acceptor, while these roles are reversed for cutting the complementary strand of opposite polarity ( 17).

In this article, this hypothesis was tested experimentally by using truncated heterodimers of BfiI that lack the DNA-binding domain from one subunit: from either the subunit carrying the inactivating H105A mutation, WT/H105A-N or from the WT subunit, H105A/WT-N. Contrary to the full length heterodimers bearing both DNA recognition domains ( Figure 3D), each truncated heterodimer has to bind DNA in a specified orientation: either the productive orientation in which the His105 residue from the WT subunit is positioned for the in-line attack on the scissile phosphate or the non-productive orientation where the residue in position for the in-line attack is the alanine from the H105A subunit ( Figure 4A and B). Hence, if the WT/H105-N heterodimer shows catalytic activity, the covalent intermediate is formed by the histidine from the full-length enzyme subunit bound to the target site on DNA, the 1° subunit. Alternatively, if the H105A/WT-N variant displays activity, the histidine nucleophile comes from the truncated subunit that is not bound to the recognition sequence, the 2° subunit. Thus, by analyzing the activities of the truncated heterodimers, we were able to identify directly which histidine residue forms the covalent intermediate during the cleavage of the bottom (3′–5′) and the top (5′–3′) DNA strands.

Contrary to the suggestion that H105 from one particular subunit of BfiI attacks the scissile bond in the bottom (3′–5′) DNA strand while the symmetry-related H105 from the opposite subunit takes this role for cutting the top (5′–3′) strand, it was found here the His from the 2° subunit not bound to the recognition site attacks sequentially the target phosphodiester bonds in both 3′–5′ and 5′–3′ strands. The equivalent histidine from the DNA-bound 1° subunit presumably acts as the proton donor/acceptor for the reactions on both strands. To match the anti-parallel polarity of the two DNA strands, the catalytic center of BfiI therefore must rotate by 180° between the two hydrolysis reactions ( Figure 4C). Thus, we demonstrate here a novel mechanism for the scission of double-stranded DNA as it requires a single active site to not only switch between strands but also to switch its orientation on the DNA.

The above reactions all contained BfiI in excess over the DNA, to favor binding of a single DNA molecule to each enzyme dimer. However, BfiI is optimally active when bound to two copies of its recognition sequence ( 18). To cut four phosphodiester bonds across two target sites, the single active site in the BfiI dimer must relocate between the scissile phosphates in the two sites, cleaving one phosphodiester bond at a time. In the synaptic complex of the BfiI dimer with two recognition sites ( Figure 4D), one subunit (B) is attached via its DNA-binding domain to recognition site X while the other subunit (A) is attached to site Y. This leaves subunit A as the 2° subunit with respect to recognition site X so H105 from A presumably forms the covalent intermediate during the sequential cutting of both strands at site X, while H105 from subunit B fulfils the proton donor/acceptor roles in both strand-scission events. Conversely, subunit B is the 2° subunit for site Y so, for cutting this second site, the H105 residues from the B and the A subunits should fulfill the same roles as those played by, respectively, the A and the B subunits when cutting site X. Hence, the two His residues may switch roles while cleaving two specific sites bound to an enzyme dimer ( Figure 4D), though it will always be the histidine from one particular subunit that attacks both bottom and top strands at each DNA site ( Figure 4C).