Genetics: Exam 3

13.3.8. What are the three basic stages of transcription? Describe what happens at each stage.

(1) Initiation: Transcription proteins assemble at the promoter to form the basal transcription apparatus and begin synthesis of RNA. (2) Elongation: RNA polymerase moves along the DNA template in a 3' to 5' direction, unwinding the DNA and synthesizing RNA in a 5' to 3' direction. (3) Termination: Synthesis of RNA is terminated, and the RNA molecule separates from the DNA template.

13.2.7. Give the names of the RNA polymerases found in eukaryotic cells and the types of RNA that they transcribe.

(1) RNA polymerase I transcribes rRNA 2. RNA polymerase II transcribes pre-mRNA, snoRNAs, and some miRNAs and snRNAs. (3) RNA polymerase III transcribes small RNA molecules such as 5S rRNA, tRNAs, and some snRNAs and miRNAs. (4) RNA polymerase IV transcribes siRNAs in plants.

12.5.18. What are some of the enzymes taking part in recombination in E. coli, and what roles do they play?

(1) RecBCD protein unwinds double-stranded DNA and can cleave polynucleotide strands. (2) RecA protein allows a single strand to invade a double-stranded DNA. (3) RuvA and RuvB proteins promote branch migration during homologous recombination. (4) RuvC protein is resolvase, a protein that resolves the Holliday structure by cleavage of the DNA. (5) DNA ligase repairs nicks or cuts in the DNA generated during recombination.

14.2.4. What are the three primary regions of mRNA sequences in bacterial cells?

(1) The 5′ untranslated region, which contains the Shine-Dalgarno sequence (2) The protein-encoding region, which begins with the start codon and ends with the stop codon (3) The 3′ untranslated region

15.1.1. What is the one gene, one enzyme hypothesis? Why was this hypothesis an important advance in our understanding of genetics?

-it was the first hypothesis to link each gene with a single protein. The one gene, one enzyme hypothesis proposed by Beadle and Tatum states that each gene encodes a single, separate protein. Now that we know more about the nature of enzymes and genes, it has been modified to the one gene, one polypeptide hypothesis because many enzymes consist of multiple polypeptides. The original hypothesis helped establish a linear link between genes (DNA) and proteins

13.2.18. The following sequence of nucleotides is found in a single-stranded DNA template: ATTGCCAGATCATCCCAATAGAT Assume that RNA polymerase proceeds along with this template from left to right. a. Which end of the DNA template is 5′, and which end is 3′? b. Give the sequence and identify the 5′ and 3′ ends of the RNA transcribed from this template.

3'- ATTGCCAGATCATCCCAATAGAT-5' RNA 3′-AUCUAUUGGGUAGAUCUGGCAAU-5′

13.2.16. The following diagram represents DNA that is part of the RNA-coding sequence of a transcription unit. The bottom strand is the template strand. Give the sequence found on the RNA molecule transcribed from this DNA, and identify the 5' and 3' ends of the RNA. 5'-ATAGGCGATGCCA-3' 3'- TATCCGCTACGGT-5' --> Template SHOW ANSWER

5'- AUAGGCGAUGCCA

7.3.11. What does the interference tell us about the effect of one crossover on another?

A positive interference value results when the actual number of double crossovers observed is less than the number of double crossovers expected from the single crossover frequencies. Thus positive interference indicates that a crossover inhibits or interferes with the occurrence of a second crossover nearby. Conversely, a negative interference value, where more double crossovers occur than expected, suggests that a crossover event can stimulate additional crossover events in the same region of the chromosome.

7.2.28. A series of two-point crosses were carried out among seven loci (a,b,c,d,e,f, and g), producing the following recombination frequencies. Map the seven loci, showing their linkage groups, the order of the loci in each linkage group, and the distances between the loci of each group. SHOW ANSWER

A--G-----D (A-G = 4 map units G-D = 8 map units) C-----B------------------E (C-B = 10 map units B-E = 18 map units) Gene F is unlinked to either of these groups; it is on a third linkage group

13.2.6. Describe the structure of the bacterial RNA polymerase holoenzyme.

Bacterial RNA polymerase consists of several polypeptide. The core enzyme for most bacterial RNA polymerase consists of five polypeptide Bacterial RNA polymerase consists of several polypeptide. The core enzyme for most bacterial RNA polymerase consists of five polypeptide subunits: two copies of the alpha subunit, and single copies of the beta, beta prime subunit, and the omega subunits. The addition of a sigma factor to the core enzyme forms the RNA polymerase holoenzyme: two copies of the alpha subunit, and single copies of the beta, beta prime subunit, and the omega subunits. The addition of a sigma factor to the core enzyme forms the RNA polymerase holoenzyme

7.3.33. Priscilla Lane and Margaret Green studied the linkage relations of three genes affecting coat color in mice: mahogany (mg), agouti (a), and ragged (Rg). They carried out a series of three-point crosses, mating mice that were heterozygous at all three loci with mice that were homozygous for the recessive alleles at these loci (P. W. Lane and M. C. Green. 1960. Journal of Heredity 51:228-230). The following table lists the progeny of the testcrosses: Note: + represents a wild-type allele. Determine the order of the loci that encode mahogany, agouti, and ragged on the chromosome; the map distances between them; and the coefficient of coincidence and the interference among these genes. Draw a picture of the two chromosomes in the triply heterozygous mice used in the testcrosses, indicating which of the alleles are present on each chromosome.

Based on the number of progeny, the non-recombinants are a + mg and + Ra + . The double crossovers are a Ra + and + + mg. These differ in the a locus; therefore, a (agouti ) is the middle locus. The recombinants between a and mg are a and + or + and mg (ignoring the Ra locus): RF= (1+1+15+9)/223 = 0.119= 0.12= 12% or 12 a.m.u The recombinants between a and Ra are + + or a Ra(Ignoring the mg locus): RF = (1+1+16+36)/ 223 = 0.25 = 25% or 25 a.m.u The C.O.C = observed dcos / expected dcos= 2/(0.12)(0.25)(213) = 2/6.4 = 0.31 Interference=1-C.O.C = 1- 0.31 = 0.69

12.4.14. In what ways is eukaryotic replication similar to bacterial replication, and in what ways is it different?

Both eukaryotic and bacterial replication of DNA replication share some basic principles: (1) Semi-conservative replication. (2) Replication origins serve as starting points for replication. (3) Short segments of RNA called primers provide a 3'-OH for DNA polymerases to begin synthesis of the new strands. (4) Synthesis occurs in a 5' to 3' direction. (5) The template strand is read in a 3' to 5' direction. (6) Deoxyribonucleoside triphosphates are the substrates. (7) Replication is continuous on the leading strand and discontinuous on the lagging strand. Eukaryotic DNA replication differs from bacterial replication in that: (1) It has multiple origins of replications per chromosome. (2) It has several different DNA polymerases with different functions. (3) Immediately following DNA replication, assembly of nucleosomes takes place

13.1.2. What are the major classes of cellular RNA?

Cellular RNA molecules are made up of six classes:(1)Ribosomal RNA, or rRNA, is found in the cytoplasm.(2)Transfer RNA, or tRNA, is found in the cytoplasm.(3)Messenger RNA is found in the cytoplasm (however, pre-mRNA is found only in the nucleus). (4)Small nuclear RNA, or snRNA, is found in the nucleus as part of riboproteins called snrps. (5)Small nucleolar RNA, snoRNA, is found in the nucleus. (6)Small cytoplasmic RNA, or scRNA, is found in the cytoplasm.

7.2.3. What effect does crossing over have on linkage?

Crossing over generates recombination between genes located on the same chromosome, and thus renders linkage incomplete.

7.2.4. Why is the frequency of recombinant gametes always half the frequency of crossing over?

Crossing over occurs at the four-strand stage, when two homologous chromosomes, each consisting of a pair of sister chromatids, are paired. Each crossover involves just two of the four strands and generates two recombinant strands. The remaining two strands that were not involved in the crossover generate two nonrecombinant strands. Therefore, the frequency of recombinant gametes is always half the frequency of crossovers.

12.3.9. Why is DNA gyrase necessary for replication?

DNA synthesis relies on single-stranded template; thus, double-stranded DNA molecules must be unwound prior to replication. During DNA unwinding by DNA helicase, tension builds up ahead of the separation (supercoiling). DNA gyrase (also referred to as topoisomerase) reduces supercoiling (relaxes tension) which builds up during DNA unwinding, preventing DNA breakage.

12.2.3. Draw a molecule of DNA undergoing theta replication. On your drawing, identify (a) the origin of replication, (b) polarity (5′ and 3′ ends) of all template strands and newly synthesized strands, (c) leading and lagging strands, (d) Okazaki fragments, and (e) locations of primers.

DRAW

12.2.4. Draw a molecule of DNA undergoing rolling-circle replication. On your drawing, identify the (a) origin, (b) template and newly synthesized strands, and (c) 5′ and 3′ ends of template and newly synthesized strands.

DRAW

12.2.5. Draw a molecule of DNA undergoing linear eukaryotic replication. On your drawing, identify (a) origin of replication, (b) polarity (5′ and 3′ ends) of all template and newly synthesized strands, (c) leading and lagging strands, (d) Okazaki fragments, and (e) locations of primers.

DRAW

7.3.10. Explain how to determine, using the numbers of progeny from a three-point cross, which of three linked loci is the middle locus.

Double crossovers always result in switching the middle gene with respect to the two nonrecombinant chromosomes. Hence, one can compare the two double crossover phenotypes with the two nonrecombinant phenotypes and see which gene is reversed. In the diagram on the facing page we see that the coupling relationship of the middle gene is flipped in the double crossovers with respect to the genes on either side. So whichever gene on the double crossover can be altered to make the double crossover resemble a nonrecombinant chromosome is the middle gene. If we take either of the double crossover products l M r or L m R, changing the M gene will make it resemble a nonrecombinant. l m r l M r

13.3.23. Most RNA molecules have three phosphate groups at the 5′ end, but DNA molecules never do. Explain this difference.

During initiation of DNA replication, DNA nucleoside triphosphates must be attached to a 3′-OH of a RNA molecule by DNA polymerase. This process removes the terminal two phosphates of the nucleotides. If the RNA molecule is subsequently removed, then a single phosphate would remain at the 5′ end of the DNA molecule. RNA polymerase does not require the 3′-OH to initiate synthesis of RNA molecules. Therefore, the 5′ end of a RNA molecule will retain all three of the phosphates from the original nucleotide triphosphate substrate.

12.2.2. How did Meselson and Stahl demonstrate that replication in E. coli takes place in a semiconservative manner?

E. coli cells grown in a medium with 15 Nitrogen isotope; the isotope integrated in the DNA. The cells then were switched to a medium with common 14Nitrogen. Samples were placed in a centrifuge. The 15Nitrogen (heavier isotope), sunk lower than 14N. DNA from cells grown in the 15N medium produced one band at the expected position; after one round of replication in the 14 N medium, after centrifuging, the band was located between the DNA 15N band and that of DNA 14N band. After 2 rounds replication, 2 bands of DNA were present. One band was located at an intermediate position between that of DNA 15N and DNA N14, and the other band was located at the expected DNA N14 position. The results were consistent with semiconservative replication.

12.3.10. What similarities and differences exist in the enzymatic activities of DNA polymerases I and III? What is the function of each DNA polymerase in bacterial cells?

Each of the five DNA polymerases has a 5′ to 3′ polymerase activity. They differ in their exonuclease activities. DNA polymerase I has a 3′ to 5′ as well as a 5′ to 3′ exonuclease activity. DNA polymerase III has only a 3' to 5' exonuclease activity.(1) DNA polymerase I carries out proofreading. It also removes and replaces the RNA primers used to initiate DNA synthesis.(2) DNA polymerase III is the primary replication enzyme and also has a proofreading function in replication Why is primase required for replication? Primase is a DNA-dependent RNA polymerase. Primase synthesizes the short RNA molecules, or primers, that have a free 3'-OH to which DNA polymerase can attach deoxyribonucleotides in replication initiation. The DNA polymerases require a free 3'-OH to which they add nucleotides, and therefore they cannot themselves initiate replication. Primase does not have this requirement.

7.4.40. A panel of cell lines was created by human-mouse somatic-cell hybridization. Each cell line was examined for the presence of human chromosomes and for the production of three enzymes. The following results were obtained: On the basis of these results, give the chromosomal locations of the genes encoding enzyme 1, enzyme 2, and enzyme 3.

Enzyme 1 is located on chromosome 9. Chromosome 9 is the only chromosome that is present in the cell lines that produce enzyme 1 and absent in the cell lines that do not produce enzyme 1. Enzyme 2 is located on chromosome 4. Chromosome 4 is the only chromosome that is present in cell lines that produce enzyme 2 (C and D) and absent in cell lines that do not produce enzyme 2 (A and B) Enzyme 3 is located on the X chromosome. The X chromosome is the only chromosome present in the three cell lines that produce enzyme 3 and absent in the cell line that does not produce enzyme 3.

13.4.13. How is transcription different in bacteria and eukaryotes? How is it similar?

Eukaryotes - three RNA polymerases each recognizes and transcribes different types of promoters, many protein transcription factors are needed to bind to the different promoters for initiation; RNA mol requires extensive post-transitional processing (adding 5' cap and 3' poly (A) tail, and removal of introns) for RNA to be functional.Bacteria- promoters tend to be more uniform in composition and only on RNA pol does transcription, RNAs are usually functional once transcription has taken place

13.4.11. Compare the roles of general transcription factors that bind to the core promoter and other transcription factors that bind to the regulatory promoter and enhancers.

General transcription factors form the basal transcription apparatus together with RNA polymerase and are needed to initiate minimal levels of transcription. Transcriptional activator proteins bring about higher levels of transcription by stimulating the assembly of the basal transcriptional apparatus at the start site.

7.1.1. What does the term recombination mean? What are two causes of recombination?

Generation of new combinations of alleles that are different from the parental combinations. independent assortment and crossing over between homologous chromosomes during meiosis

7.2.5. What is the difference between genes in coupling configuration and genes in repulsion? How does the arrangement of linked genes (whether they are in coupling or repulsion) affect the results of a genetic cross?

Genes in coupling configuration have two wild-type alleles on the same chromosome and the two mutant alleles on the homologous chromosome. Genes in repulsion have a wild-type allele of one gene together with the mutant allele of the second gene on the same chromosome, and vice versa on the homologous chromosome. The two arrangements have opposite effects on the results of a cross. For genes in coupling configuration, most of the progeny will be either wild-type for both genes, or mutant for both genes, with relatively few that are wild-type for one gene and mutant for the other. For genes in repulsion, most of the progeny will be mutant for only one gene and wild-type for the other, with relatively few recombinants that are wild-type for both or mutant for both.

7.2.18. In cucumbers, heart-shaped leaves (hl) are recessive to normal leaves (Hl), and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and for number of spines are located on the same chromosome; findings from mapping experiments indicate that they are 32.6 m.u. apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F1 are crossed with plants that have heart-shaped leaves and numerous spines. What phenotypes proportions are expected in the progeny of this cross?

Heart-shaped, numerous spines = 33.7% Normal, few spines = 33.7% Heart-shaped, few spines = 16.3% Normal, numerous spines = 16.3%

12.3.24. The following diagram represents a DNA molecule that is undergoing replication. Draw in the strands of newly synthesized DNA and identify (a) the polarity of the newly synthesized strands, (b) leading and lagging strands, (c)Okazaki fragments, and (d) RNA primers. SHOW ANSWER

IMAGE

14.2.29. How would the deletion of the Shine-Dalgarno sequence affect a bacterial mRNA?

In bacteria, the small ribosomal subunit binds to the Shine-Dalgarno sequence to initiate translation. If the Shine-Dalgarno sequence is deleted, then translation initiation cannot take place, preventing protein synthesis.

12.4.15. What is the end-replication problem? Why, in the absence of telomerase, do the ends of linear chromosomes get progressively shorter each time the DNA is replicated?

In each replication, the DNA molecules in a chromosome would become slightly shorter. -- Eukaryotes solve the problem by locating highly repeated DNA sequence at the end, or telomeres, of each linear chromosome In the absence of telomerase, DNA polymerase will be unable to add nucleotides to the end of the strand. after multiple round of replication without a functional telomerase the chromosome will become progressively shorter.

12.2.20. Phosphorus is required to synthesize the deoxyribonucleoside triphosphates used in DNA replication. A geneticist grows some E. coli in a medium containing nonradioactive phosphorus for many generations. A sample of the bacteria is then transferred to a medium that contains a radioactive isotope of phosphorus (32P). Samples of the bacteria are removed immediately after the transfer and after one and two rounds of replication. Assume that newly synthesized DNA contains 32P and the original DNA contains nonradioactive phosphorus. What will be the distribution of radioactivity in the DNA of the bacteria in each sample? Will radioactivity be detected in neither, one, or both strands of the DNA?

In the initial sample removed immediately after transfer, no 32P should be incorporated into the DNA because replication in the medium containing 32P has not yet occurred. After one round of replication in the 32P-containing medium, one strand of each newly synthesized DNA molecule will contain 32P, while the other strand will contain only nonradioactive phosphorous. After two rounds of replication in the 32P-containing medium, 50% of the DNA molecules will have 32P in both strands, while the remaining 50% will contain 32P in one strand and nonradioactive phosphorous in the other strand.

14.2.7. How is the poly(A) tail added to pre-mRNA? What is the purpose of the poly(A) tail?

Initially, a complex consisting of several proteins forms on the 3′ UTR of the pre-mRNA molecule. Cleavage and polyadenylation specificity factor (CPSF) bind to the AAUAAA consensus sequence, which is located upstream of the 3′ cleavage site. Another protein, cleavage stimulation factor (CsTF), binds downstream of the cleavage site. Two cleavage factors (CFI and CFII) and polyadenylate polymerase (PAP) also become part of the complex. Once the complex has formed, the pre-mRNA is cleaved. CsTF and the two cleavage factors leave the complex. PAP adds approximately 10 adenine nucleotides to the 3′ end of the pre-mRNA molecule. The addition of the short poly(A) tail allows for the binding of the poly(A) binding protein (PABII) to the tail. PABII increases the rate of polyadenylation, which subsequently allows for more PABII protein to bind the tail. b. The presence of the poly(A) tail increases the stability of the mRNA molecule through the interaction of proteins at the poly(A) tail. The length of the poly(A) tail influences the time in which the transcript remains intact and available for translation. The poly(A) tail also assists with the binding of the ribosome to the mRNA.

14.1.2. What are some characteristics of introns?

Introns are intervening sequences, typically do not encode proteins. Eukaryotic genes commonly contain introns but in bacterial genes. The number of introns found in an organism's genome is typically related to complexity— more complex organisms = more introns

7.2.16. In 1916, Dorothy Osborn suggested that pattern baldness (the common type of baldness seen in most males) is a sex-influenced trait (see Chapter 5) that is dominant in males and recessive in females. Other research has found that pattern baldness is influenced by genes on the X chromosome. Would you expect to see independent assortment between genetic markers on the X chromosome and pattern baldness if (a) pattern baldness is sex-influenced and (b) if pattern baldness is X-linked recessive? Explain your answer. (Research has also identified variation at autosomal genes that are associated with pattern baldness.)

It depends on the distance; it will show linkage to some X-linked genetic markers but may assort independently of more distant X-linked genetic markers. (a) If pattern baldness were a sex-influenced trait, we would see independent assortment between genetic markers on the X chromosome and pattern baldness. By definition, sex-influenced traits are encoded by autosomal genes, which will assort independently of X-linked genes, because they are on different chromosomes. (b) If pattern baldness were X-linked recessive, it would be encoded by a gene on the X chromosome. If this gene and the X-linked markers were close together on the X chromosome, they would not assort independently. If however, the gene for pattern baldness and the X-linked markers were located far apart, so that crossing over between them occurred every meiosis, they might assort independently.

15.1.17. Compounds I, II, and III are in the following biochemical pathway: Mutation a inactivates enzyme A, mutation b inactivates enzyme B, and mutation c inactivates enzyme C. Mutants, each having one of these defects, were tested on a minimal medium to which compound I, II, or III was added. Fill in the results expected of these tests by placing a plus sign (+) for growth or a minus sign (−) for no growth in the following table. Pointing left to right the precursor is followed by an arrow labeled enzyme A, pointing to compound 1. An arrow angled downward from compound 1 is labeled enzyme B and points to compound 2. An arrow pointing right from compound 2 is labeled enzyme C and points to compound 3.

Mutant A: I: + II: + III: + Mutant B: I: - II: + III: + Mutant C: I: - II: - III: + Mutant strain a enzyme activity for A is loss but other enzymes B and C activity still remains. So in Mutant strain a compound I is added, so anyway compound I will be get for the mutant strain a. It not depends on enzyme A. It grows in all the compounds. Mutant strain b enzyme activity for B is loss but other enzymes A and C activity still remains. But if we add compound I it not able to convert into compound B. So it not able to grow in that medium. But if we add compound B it convert in to compound C by enzyme C and it will grow.

7.2.7. How would you test to see if two genes are linked?

One first obtains individuals that are heterozygous for both genes. This may be achieved by crossing an individual homozygous dominant for both genes to one homozygous recessive for both genes, resulting in a heterozygote with genes in coupling configuration. Alternatively, an individual that is homozygous recessive for one gene may be crossed to an individual homozygous recessive for the second gene, resulting in a heterozygote with genes in repulsion. Then the heterozygote is mated to a homozygous recessive tester and the progeny of each phenotypic class are tallied. If the proportion of recombinant progeny is far less than 50%, the genes are linked. If the results are not so clear-cut, then they may be tested by chi-square, first for equal segregation at each locus, then for independent assortment of the two loci. Significant deviation from results expected for independent assortment indicates linkage of the two genes.

14.2.28. Draw a typical eukaryotic gene nd the pre-mRNA and mRNA derived from it. Assume that the gene contains three exons. Identify the following items, and for each item, give a brief description of its function: a. 5' untranslated region b. Promoter c. AAUAAA consensus sequence d. Transcription start site e. 3' untranslated region f. Introns g. Exons h. Poly (A) tail I. 5' cap SHOW ANSWER

PICTURE

12.3.11. Why is primase required for replication?

Primase is a DNA-dependent RNA polymerase. Primase synthesizes the short RNA molecules, or primers, that provide a 3'- OH to which DNA polymerases require a 3'-OH to which they add nucleotides, and therefore they cannot initiate replication. Primase does not have this requirement.

13.1.1. Draw an RNA nucleotide and a DNA nucleotide, highlighting the differences. How is the structure of RNA similar to that of DNA? How is it different? (See Chapter 10 for a review of DNA structure.)

RNA and DNA are polymers of nucleotides that are held together by phosphodiester bonds. An RNA nucleotide contains ribose, whereas a DNA nucleotide contains deoxyribose. RNA contains uracil but not thymine. DNA contains thymine but not uracil. RNA is typically single stranded, even though RNA molecules can pair with other complementary sequences. DNA molecules are almost always double stranded.

14.2.11. What is RNA editing? Explain the role of guide RNAs in RNA editing.

RNA editing alters the sequence of a RNA molecule after transcription either by the insertion, deletion, or modification of nucleotides within the transcript. The guide RNAs (gRNAs) provide templates for the alteration of nucleotides in RNA molecules undergoing editing and are complementary to regions within the pre-edited RNA molecule. At the complementary regions, the gRNAs base pair to the pre-edited RNA molecule. Alteration of nucleotides at the paired region follows.

13.3.36. Elaborate repair mechanisms that prevent permanent mutations in DNA are associated with replication (see Chapter 18), yet no similar repair process is associated with transcription. Can you think of a reason for this difference between replication and transcription? (Hint: Think about the relative effects of a permanent mutation in a DNA molecule and one in an RNA molecule.)

RNA transcripts are not inherited and their influence in a cell is only transient; therefore, the cell doesn't waste energy repairing RNA

14.2.9. Explain the process of pre-mRNA splicing in nuclear genes.

Removal of an intron from the pre-mRNA requires the assembly of the spliceosome complex on the pre-mRNA, cleavage at both the 5' and 3' splices sites of the intron, and 2 trans-esterification reactions ultimately leading to the joining of the 2 exons. snRNP U1 binds to the 5' splice site through complementary base pairing of the U1 snRNA. snRNP U2 binds to the branch point within the intron. U5 and U4-U6 complex joins the spliceosome, resulting in the looping of the intron so that the ranch point and 5' splice site of the intron are now adjacent to each other. U1 and U4 dissociate from spliceosome, and spliceosome is activated. Pre-mRNA is cleaved at 5' splice site, producing an exon with a 3'-OH. 5' end of intron folds back and forms 5'-2' phosphodiester linkage—lariat— through the first trans-esterification reaction with the adenine nucleotide at the branch point of the intron. 3' splice site is cleaved and then ligated to the 3'-OH of the first exon through the second trans-esterification reaction. Thus, the exons are now joined and the intron has been excised.

14.2.5. What is the function of the Shine-Dalgarno consensus sequence?

Ribosome binding site on mRNA molecule

13.3.22. List at least five properties that DNA polymerases and RNA polymerases have in common. List at least three differences. See Chapter 12 for information on DNA polymerases.

Similarities: (1) Both use DNA templates; (2) DNA templates are read in the 3′ to 5′ direction; (3) the complementary strand is synthesized in a 5′ to 3′ direction that is antiparallel to the template; (4) both use nucleoside triphosphates as substrates; and (5) their actions are enhanced by accessory proteins. Differences: (1) RNA polymerases use ribonucleoside triphosphates as substrates, whereas DNA polymerases use deoxyribonucleoside triphosphates; (2) DNA polymerases require a primer that provides an available 3′-OH group where synthesis begins, whereas RNA polymerases do not require primers to begin synthesis; and (3) RNA polymerases synthesize a copy of only one of the DNA strands, whereas DNA polymerases can synthesize copies of both strands.

13.3.21. Write the consensus sequence for the following set of nucleotide sequences. AGGAGTT AGCTATT TCGAATA ACGAAAA TCCTAAT TGCAATT SHOW ANSWER

T/AGCAATT

14.2.6. What is the 5′ cap? How is the 5′ cap added to eukaryotic pre-mRNA? What is the function of the 5′ cap?

The 5' end of eukaryotic mRNA is modified by the addition of the 5' cap. The cap consists of an extra guanine nucleotide linked 5'to 5' to the mRNA molecule.This nucleotide is methylated at position 7 of the base. The ribose sugars of adjacent bases may be methylated at the 2' -OH -Initially, the terminal phosphate of the three 5′ phosphates linked to the end of the mRNA molecule is removed.A guanine nucleotide is attached to the 5′ end of the mRNA using a 5′ to 5′ phosphate linkage.Then, a methyl group is attached to position 7 of the guanine base. Ribose sugars of adjacent nucleotides may also be methylated, but at the 2′ -OH -CAP binding proteins recognize the 5′ cap and stimulate binding of the ribosome to the 5′ cap and to the mRNA molecule. The 5′ cap may also increase mRNA stability in the cytoplasm. Finally, the 5′ cap is needed for efficient splicing of the intron that is nearest the 5′ end of the pre-mRNA molecule.

13.3.28. The following diagram represents a transcription unit on a DNA molecule. D N A double-strand is represented by two parallel horizontal lines. The top strand runs from 5 prime and the bottom strand runs from 3 prime. The transcription start site is marked a little before the center of the top strand, using a downward arrow. a. Assume that this DNA molecule is from a bacterial cell. Draw the approximate locations of the promoter and terminator for this transcription unit. Assume that this DNA molecule is from a eukaryotic cell. Draw the approximate location of an RNA polymerase II promoter.

The RNA molecule would be complementary to the template strand, contain uracil, and be synthesized in an antiparallel fashion. The sequence would be:5′-A U A G G C G A U G C C A-3′.The RNA strand contains the same sequence as the non-template DNA strand except that the RNA strand contains uracil in place of thymine.

14.2.30. What would be the most likely effect of moving the AAUAAA consensus sequence shown in Figure 14.7 10 nucleotides upstream?

The cleavage site would also be moved 10 nucleotides upstream, likely resulting in shorter mRNA.

7.2.22. Recombination frequencies between three loci in corn are shown in the following table: R and W2 = 17% R and L2 = 35% W2 and L2 = 18% SHOW ANSWER

The distances between the genes are indicated by the recombination rates. Because R and L2 have the highest recombination rate, they must be the farthest apart, and W2 is in the middle. The order of the genes is R, W2, and L2

7.2.9. Why do calculated recombination frequencies between pairs of loci that are located far apart underestimate the true genetic distances between loci?

The further apart two loci are, the more likely it is to get double crossovers between them. Unless there are marker genes between the loci, such double crossovers will be undetected because double crossovers give the same phenotypes as nonrecombinants. The calculated recombination frequency will underestimate the true crossover frequency because the double crossover progeny are not counted as recombinants.

14.1.22. Duchenne muscular dystrophy is caused by a mutation in a gene that comprises 25 million base pairs and specifies a protein called dystrophin. However, less than 1% of the gene actually encodes the amino acids in the dystrophin protein. On the basis of what you now know about gene structure and RNA processing in eukaryotic cells, provide a possible explanation for the large size of the dystrophin gene. SHOW ANSWER

The large size is probably due to the presence of many intervening sequences, or introns, within the coding region of the gene

7.4.41. The locations of six deletions have been mapped to a Drosophila chromosome, as shown in the following deletion map. Recessive mutations, a, b, c, d, e, and f are known to be located in the same region as the deletions, but the order of the mutations on the chromosome is not known. When flies homozygous for the recessive mutations are crossed with flies heterozygous for the deletions, the following results are obtained, in which "m" represents a mutant phenotype and a plus sign (+) represents the wild type. On the basis of these data, determine the relative order of the six mutant genes on the chromosome. SHOW ANSWER

The location of f is ambiguous; it could be in either location shown in the deletion map. Deletion 1 --> Deletion 2 --> Deletion 3 --> Deletion 4 (____c) --> Deletion 5 (___e)--> Deletion 6 (_d_____b)

15.1.16. Sydney Brenner isolated Salmonella typhimurium mutants that were implicated in the biosynthesis of tryptophan and would not grow on a minimal medium. When this bacterial mutant was tested on a minimal medium to which one of the four compounds (indole glycerol phosphate, indole, anthranilic acid, and tryptophan) had been added, the growth responses shown in the following table were obtained. Give the order of indole glycerol phosphate, indole, anthranilic acid, and tryptophan in a biochemical pathway leading to the synthesis of tryptophan. Indicate which step in the pathway is affected by each of the mutations. SHOW ANSWER

The mutations can be assembled into four groups: Group 1 mutants (trp-1, trp-10, trp-11, trp-9, trp-6, and trp-7) can grow only a minimal medium supplemented with tryptophan. Group 2 mutants (trp-3) can grow on a minimal medium supplemented with either tryptophan or indole. Group 3 mutants (trp-2 and trp-4) can grow on a minimal medium supplemented with tryptophan, indole, or indole glycerol phosphate. Group 4 mutants (trp-8) can grow on a minimal medium supplemented with the addition of tryptophan, indole, indole glycerol phosphate, or anthranilic acid Precursor (Group 4) --> anthranilic acid (Group 3)--> indole glycerol phosphate (Group 2) --> indole (Group 1) --> tryptophan

7.2.20. Alleles A and a reside at a locus on the same chromosome as a locus with alleles B and b. Aa Bb is crossed with aa bb, and the following progeny are produced: Aa Bb 5 Aa bb 45 aa Bb 45 aa bb 5 What conclusion can be drawn about the arrangement of the genes on the chromosome in the Aa Bb parent?

The results of this testcross reveal that Aa bb and aa Bb, with far greater numbers, are the progeny that received nonrecombinant chromatids from the Aa Bb parent. Given that all the progeny received ab from the aa bb parent, the nonrecombinant progeny received either an Ab or an aB chromatid from the Aa Bb parent. Therefore, the A and B loci are in repulsion in the Aa Bb parent. Aa Bb and aa bb are the recombinant classes, and their frequencies indicate that the genes A and B are 10 m.u. apart

14.2.8. What makes up the spliceosome? What is the function of the spliceosome?

The spliceosome consists of five small ribonucleoproteins (snRNPs). Each snRNP is composed of multiple proteins and a single small nuclear RNA molecule or snRNA. The snRNPs are identified by which snRNA (U1, U2, U3, U4, U5, or U6) each contains. Splicing of pre-mRNA nuclear introns takes place within the spliceosome

12.2.7. What substrates are used in DNA synthesis?

The substrates used for DNA synthesis are the four types of deoxyribonucleoside triphosphates: (A, T, C, G) deoxyadenosine triphosphate, deoxyguanosine triphosphate, deoxycytosine triphosphate, and deoxythymidine triphosphate

13.3.10. What are the two basic types of terminators found in bacterial cells? Describe the structure of each type.

The two basic types of terminators in bacterial cells are rho-independent and rho-dependent terminators. Rho-independent terminators consist of inverted repeats that can form a hairpin structure. Immediately following the inverted repeats is a string of six adenine nucleotides. Rho-dependent terminators require the interaction of the protein rho with RNA polymerase. Two features are typical for rho-dependent termination: (1) variable DNA sequences that cause RNA polymerase to pause during transcription and (2) upstream from variable region lies DNA sequence that encodes RNA devoid of secondary structure, which also serves as the rho binding site.

13.2.4. What parts of DNA make up a transcription unit? Draw a typical bacterial transcription unit, and identify its parts.

The typical bacterial promoter consists of the -35 (TTGACA) and -10 (TATAAT)consensus sequences. An upstream element rich in AT sequences is found only in some bacterial promoters and is located upstream of the -35 consensus sequence typically around -40 or -60.Transcription start site 5′—TTGACA----------------------TATAAT-------3′3′—AACTGT----------------------ATATTA-------5′-35 -10 Region Region

7.2.23. In tomatoes, dwarf (d) is recessive to tall (D), and opaque (light-green) leaves (op) are recessive to green leaves (Op). The loci that determine the height and leaf color are linked and separated by a distance of 7 m.u. For each of the following crosses, determine the phenotypes and proportions of progeny produced. a. DOpdop×dopdOp b. DopdOp×dopdop c. DOpdop×DOpdop d. DopdOp×DopdOp

a). D Op / d op X d op / d op -Parents Distance between genes (m.u.) = Recombination frequency (%) D Op / d op genotype produces 4 types of gametes as below: Recombinant gametes (7%): D op = 3.5% d Op= 3.5% Non-recombinant gametes (93%): D Op= 46.5% d OP =46.5% Genotype, d op / d op produces only one type of gametes, d op. D OP/d op (46.5%) = Tall, green D op/d op (3.5%) = Tall, opaque d OP/d op (3.5%)= Dwarf, green d op/d op (46.5%) = Dwarf, opaque b). D op / d Op X d op / d op -Parents Distance between genes (m.u.) = Recombination frequency (%) D op / d Op genotype produces 4 types of gametes as below: Recombinant gametes (7%): D Op = 3.5% d op= 3.5% Non-recombinant gametes (93%): D op= 46.5% d OP =46.5% Genotype, d op / d op produces only one type of gametes, d op. D OP/d op (3.5%) = Tall, green D op/d op (46.5%) = Tall, opaque d OP/d op (46.5%)= Dwarf, green d op/d op (3.5%) = Dwarf, opaque c). D Op / d op X D Op / d op -----Parents D Op / d op genotype produces 4 types of gametes as below: Recombinant gametes (7%): D op = 3.5% = 0.035 d Op= 3.5%=0.035 Non-recombinant gametes (93%): D Op= 46.5%=0.465 d OP =46.5%=0.465 Tall, green 0.716225 or 71.6% Tall, opaque 0.033775 = 3.4% Dwarf, green 0.033775 = 3.4% Dwarf, opaque 0.216225 = 21.6% d). D op / d Op X D op / d Op -Parents Distance between genes (m.u.) = Recombination frequency (%) D op / d Op genotype produces 4 types of gametes as below: Recombinant gametes (7%): D Op = 3.5% d op= 3.5% Non-recombinant gametes (93%): D op= 46.5% d OP =46.5% Tall, green 0.501225 or 50.1% Tall, opaque 0.248775 or 24.9% Dwarf, green 0.248775 or 24.9% Dwarf, opaque 0.001225 or 0.1%

7.2.15. In silkmoths ( Bombyx mori), red eyes (re) and white-banded wings (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome, A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have wild-type eyes and wild-type wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are: wild-type eyes, wild-type wings = 418 red eyes, wild-type wings = 19 wild-type eyes, white-banded wings = 16 red eyes, white-banded wings = 426 a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes? b. What is the rate of recombination between the gene for red eyes and the gene for white-banded wings? SHOW ANSWER

a. 1/4 wild-type eyes, wild-type wings; 1/4 red eyes, wild-type wings; 1/4 wild-type eyes, white-banded wings; and 1/4 red eyes, white-banded wings b. The F1 heterozygotes inherited a chromosome with alleles for red eyes and white-banded wings (re wb) from one parent and a chromosome with alleles for wild-type eyes and wild-type wings (re+wb+) from the other parent. These are, therefore, the phenotypes of the nonrecombinant progeny, present in the highest numbers. The recombinants are the 19 with red eyes, wild-type wings and the 16 with wild-type eyes, white-banded wings. --> recombination frequency = (19+16)/879 x 100%= 4.0% The distance between the genes is 4 map units.

14.2.12. Summarize the different types of processing that can take place in pre-mRNA.

a. Addition of 5' cap to 5' end of pre-mRNA b. Cleavage 3' end of site downstream of the AAUAAA consensus sequence of the last exon c. Addition of poly(A) tail to 3' end of mRNA after cleavage sited. Removal of introns (splicing)

7.2.2. In a testcross for two genes, what types of gametes are produced with (a) complete linkage, (b) independent assortment, and (c) incomplete linkage?

a. Complete linkage of two genes means that only nonrecombinant gametes will be produced; the recombination frequency is zero. b.Independent assortment of two genes will result in 50% of the gametes being recombinant and 50% being nonrecombinant, as would be observed for genes on two different chromosomes. Independent assortment may also be observed for genes on the same chromosome, if they are far enough apart that one or more crossovers occur between them in every meiosis. c. Incomplete linkage means that greater than 50% of the gametes produced are nonrecombinant and less than 50% of the gametes are recombinant; the recombination frequency is greater than 0 and less than 50%.

7.3.34. Fine spines (s), smooth fruit (tu), and uniform fruit color (u) are three recessive traits in cucumbers, the genes for which are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross, and the following progeny are produced by this testcross: Determine the order of these genes on the chromosome. Calculate the map distances between the genes. Determine the coefficient of coincidence and the interference among these genes. List the genes found on each chromosome in the parents used in the testcross.

a. For determining the order of genes we first need to find the non recombinants and the double crossovers. In this case: s u Tu and S U tu are non recombinants while s u tu and S U Tu are double crossovers. We simple look at the different gene. In this case it is Tu which means that the middle spot belongs to Tu. b. We now calculate the map distance: S Tu and s tu are recombinants so the recombinant frequency will be equal to = 2+4+21+21= 48/230 The answer comes out to be 20.9%. Hence the map distance is 20 map units. Now coming to U Tu we have u tu and U Tu as the recombinants. The recombinant frequency will be equal to = 2+4+13+17= 36/230. The answer comes out to be 15.6% or 15.6 map units. Hence the map distance between the given genes is 15.6 map units. C. Now coming to coefficient of incidence and interference. First finding the expected dco frequency which is the product of RF1 X RF2. This can be calculated as: = (15.6%)(20.9%) = 3.26% The observed dco frequency can be calculated as the homozygous genotype/total number of progeny. This can be calculated as: 6/230= 2.6% Finally the coefficient of coincidence= observed dcos/ expected dcos. This can be calculated as: 2.6/3.26 = 0.8 Interference can be calculated as: 1- Coefficient of coincidence= 1-0.8 = 0.2 d. The chromosomes of parents will look as following: Test cross parent: s tu u and s tu u For heterozygous parent: S tu U and s Tu u.

13.2.19. Assume that a mutation occurs in the gene that encodes each of the following RNA polymerases. Match the mutation with its possible effects by placing the correct letter or letters in the blanks below. There may be more than one effect for each mutated polymerase. A mutation in the gene that codes for Effects RNA polymerase I RNA polymerase II RNA polymerase III The possible effects of a.tRNA are not synthesized. b. Some ribosomal RNA is not synthesized. c. Ribosomal RNA is not processed. d. pre-mRNA is not processed. e. Some mRNA molecules are not degraded. f. pre-mRNA is not synthesized.

a. III b. I c. I d. II e. III f. II

12.3.26. What would be the effect on DNA replication of mutations that destroyed each of the following activities of DNA polymerase I? a. 3' --> 5' exonuclease activity b. 5' --> 3' exonuclease activity c. 5' --> 3' polymerase activity SHOW ANSWER

a. More errors in replication b. Primers would not be removed c. Primers that had been removed would not be replaced

13.1.15. An RNA molecule has the following percentages of bases: A=23%, U=42%, C = 21%, and G=14% a. Is this RNA single-stranded or double-stranded? How can you tell? b. What would be the percentages of bases in the template strand of the DNA that contains the gene for this RNA? SHOW ANSWER

a. Single-stranded. If it was double-stranded, we would expect nearly equal percentages of adenine and uracil, as well as equal percentages of guanine and cytosine. In this RNA molecule, the percentages of these bases are not equal. b. A=42%, T=23%, C=14%, and G= 21%

7.3.32. Waxy endosperm (wx), shrunken endosperm (sh), and yellow seedlings (v) are encoded by three recessive genes in corn that are linked on chromosome 5. A corn plant homozygous for all three recessive alleles is crossed with a plant homozygous for all the dominant alleles. The resulting F1 is then crossed with a plant homozygous for the recessive alleles in a three-point testcross. The progeny of the test cross are: a. Determine the order of these genes on the chromosome. b. Calculate the map distances between the genes. c. Determine the coefficient of coincidence and the interference among these genes. SHOW ANSWER

a. The gene order is Wx, V, Sh b. Wx-V distance is 7%, or 7 map units sh-V distance is 30%, or 30 map units c. Coefficient of coincidence = 0.80 Interference = 0.20

7.2.19. In tomatoes, tall (D) is dominant over dwarf (d), and smooth fruit (P) is dominant over pubescent fruit (p), which is covered with fine hairs. A farmer has two tall and smooth-fruit tomato plants, which we will call plant A and plant B. The farmer crosses plants A and B with the same dwarf and pubescent-fruit plant and obtains the following numbers of progeny: Dd Pp 122;2 Dd pp 6;82 dd Pp 4;82 dd pp 124;4 a. What are the genotypes of plant A and plant B? b. Are the loci that determine the height of the plant and pubescence linked? If so, what is the rate of recombination between them? c. Explain why different proportions of progeny are produced when plant A and plant B are crossed with the same dwarf and pubescent-fruit plant.

a. The genotypes of both plants are DdPp. b. Yes. From the cross of plant A, the map distance is 10/256 = 3.9% or 3.9 m.u. The cross of plant B gives 6/170 = 3.5% or 3.5 m.u. If we pool the data from the two crosses, we get 16/426 = 3.8% or 3.8 m.u. c. The two plants have different coupling configurations. In plant A, the dominant alleles D and P are coupled; one chromosome is D P and the other is d p. In plant B, they are in repulsion; its chromosomes have D p and d P.

7.2.25. In D. melanogaster, ebony body (e) and rough eyes (ro) are encoded by autosomal recessive genes found on chromosome 3; they are separated by 20 m.u. The gene that encodes forked bristles (f) is X-linked recessive and assorts independently of e and ro. Give the phenotypes of progeny and their expected proportions when a female of each of the following genotypes is test-crossed with a male. a. e+ro+erof+f b. e+roero+f+f

a. The recombination frequency between e and ro is 20% as they are separated on chromosome 3 by 20 map units.So each of the recombinants that is e+ ro and e ro+ will be 10% and each of the nonrecombinants that is e+ ro+ and e ro will be 40%.Since f sorts independently,so each of these will then be split equally among f+ and f. Phenotypes of the progeny and their expected proportion:e+ ro+ f+ 20%e+ ro+ f 20%e ro f+ 20%e ro f 20%e+ ro f+ 5%e+ ro f 5%e ro+ f+ 5%e ro+ f 5% b. The calculations are same as in a). but only here the recombinants are e+ ro+ and e ro and the nonrecombinants are e+ ro and e ro+.Phenotypes of the progeny and their expected proportion: e+ ro+ f+ 5% e+ ro+ f 5% e ro f+ 5% e ro f 5% e+ ro f+ 20% e+ ro f 20% e ro+ f+ 20% e ro+ f 20%

7.2.14. In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). a. What will the results of the testcross be if the loci that control banding and color are linked with no crossing over? b. What will the results of the testcross be if the loci assort independently (see Chapter 3)? c. What will the results of the testcross be if the loci are linked and 20 m.u. apart?

a.With absolute linkage, there will be no recombinant progeny. The F1 inherited banded and yellow alleles (BBCY) together on one chromosome from the banded yellow parent and unbanded and brown alleles (BOCBw) together on the homologous chromosome from the unbanded brown parent. Without recombination, all the F1 gametes will contain only these two allelic combinations, in equal proportions. Therefore, the F2 testcross progeny will be ½ banded, yellow and ½ unbanded, brown. b. With independent assortment, the progeny will be:¼ banded, yellow¼ banded, brown¼ unbanded, yellow¼ unbanded, brown c. The recombination frequency is 20%, so each of the two classes of recombinant progeny must be 10%. The recombinants are banded, brown and unbanded, yellow. The two classes of nonrecombinants are 80% of the progeny, so each must be 40%. The nonrecombinants are banded, yellow and unbanded, brown. In summary: 40% banded, yellow 40% unbanded, brown 10% banded, brown 10% unbanded, yellow

13.3.9. Draw a typical bacterial promoter, and identify any common consensus sequences.

promoter, transcription start site, RNA coding region, terminator (pic on other quizlet) The typical bacterial promoter consists of the -35 (TTGACA) and -10 (TATAAT)consensus sequences. An upstream element rich in AT sequences is found only in some bacterial promoters and is located upstream of the -35 consensus sequence typically around -40 or -60.Transcription start site 5′—TTGACA----------------------TATAAT-------3′3′—AACTGT----------------------ATATTA-------5′-35 -10 Region Region

7.2.17. A geneticist discovers a new mutation in D. melanogaster that causes the flies to shake and quiver. She calls this mutation quiver (qu) and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding the quiver phenotype is linked to the recessive gene encoding vestigial (reduced) wings (vg). She crosses a fly homozygous for the quiver and vestigial traits with a fly homozygous for the wild-type traits and then uses the resulting F1 females in a testcross. She obtains the following flies from this test cross: vg+qu+ = 230 vgqu = 224 vgqu+ = 97 vg+qu = 99 total = 650 Are the genes that cause vestigial wings and the quiver phenotype linked? Do a chi-square test of independence to determine whether the genes have assorted independently. SHOW ANSWER

the genes are linked and have not assorted independently To test for independent assortment, we first test for equal segregation at each locus, then test whether the two loci sort independently.Test for vg: Observed vg = 224 + 97 = 321 Observed vg+ = 230 + 99 = 329 Expected vg or vg+ = ½ × 650 = 325 (321 - 325)2/325 + (329 325)2/325 = 16/325 + 16/325 = 0.098 We have n -1 degrees of freedom, where n is the number of phenotypic classes = 2, so just 1 degree of freedom. From Table 3.4, we see that the P value is between 0.7 and 0.8. So these results do not deviate significantly from the expected 1:1 segregation.Similarly, testing for sps,we observe 327 sps+and 323 sps and expect ½ × 650 = 325 of each: x2 = 4/325 + 4/325 = .025, again with 1 degree of freedom. The P value is between 0.8 and 0.9, so these results do not deviate significantly from the expected 1:1 ratio. Finally, we test for independent assortment, where we expect 1:1:1:1 phenotypic ratios, or 162.5 of each. We have four phenotypic classes, giving us three degrees of freedom. The chi-square value of 102.5 is off the chart, so we reject independent assortment.Instead, the genes are linked, and the RF = (97 + 99)/650 100% = 30%, giving us 30 map units between them.

12.2.1. What is semiconservative replication?

two strands of DNA unzip, and a new strand is assembled onto each 'conserved' strand. The replicated double helix consists of one old strand and one newly synthesized strand -semiconservative replication, the original two strands of the double helix serve as templates for new strands of DNA. When replication is complete, two double stranded DNA molecules will be present. Each will consist of one original template strand and one newly synthesized strand that is complementary to the template