BIO 320 quiz 3

O6. Deamination of adenine creates a base called hypoxanthine which prefers to base pair with cytosine. If the adenine in the DNA sequence shown below is deaminated, what will be the sequence of the resulting mutant DNA molecule? 5' CGGTC 3' 3' GCCAG 5' A. 5' CGGCC 3' 3' GCCGG 5' B. 5' CGGGC 3' 3' GCCCG 5' C. 5' GCCAG3' 3'CGGTC 5' D. 5' CGGAC 3' 3' GCCTG5'

A. 5' CGGCC 3' 3' GCCGG 5'

O22. Hydroxylamine is a mutagen that adds hydroxyl groups to cytosine, thereby enabling the modified cytosine to basepair with adenine. If the following sequence of DNA is treated with hydroxylamine, what mutant sequence will result after DNA replication? 5'-ATGT-3' 3'-TACA-5' A. 5'-ATAT-3' 3'-TATA-5' B. 5'-ATTT-3' 3'-TAAA-5' C. 5'-TATA-3' 3'-ATAT-5' D. 5'-ATCT-3' 3'-TAGA-5' E. 5'-ATGT-3' 3'-TACA-5'

A. 5'-ATAT-3' 3'-TATA-5'

M7. Assume that a replication fork starts at the left of the DNA sequence shown below and moves towards the right. What is the sequence the DNA synthesized as the leading strand? 5'GGTATTCATCT3' 'CCATAAGTAGA5' A. 5'GGTATTCATCT3' B. 5'TCTACTTATGG3' C. 5'AGATGAATACC3' D. 5'CCATAAGTAGT3' E. 3'CCATAAGTAGA5'

A. 5'GGTATTCATCT3'

C21. An inversion mutation inverts the entire transcribed part of a gene, relative to the promoter, as shown in the figure. If the transcript from a segment of the unmutated gene has the sequence 5'CAUCGGUGCCA3, what is the sequence of the transcript (if any) from the same segment of the inverted gene? A. 5'UGGCACCGAUG3' B. 5'CAUCGGUGCCA3' C. 5'ACCGUGGCUAC3' D. 5'CUAGCCACGGU3' E. The inverted gene will not be transcribed.

A. 5'UGGCACCGAUG3'

O17. A his- strain of E. coli contains a mutation that substitutes a GC basepair for an AT basepair present in the wild-type his+ allele. Which mutagen would be best able to detect reversion of this mutant to his+? A. A base modifier that deaminates cytosine, converting it to uracil. B. Ionizing radiation leading to chromosome breakage. C. An intercalating chemical mutagen such as ethidium bromide or acridine orange. D. Any of these types of mutagens could be detected equally efficiently. E. Insertion mutations due to transposon activity.

A. A base modifier that deaminates cytosine, converting it to uracil.

B18. A single-base transition mutation in a bacterial gene changes a TGG codon to TGA. Which of the following mutations is most likely to be able to suppress the first mutation? A. A transition mutation in a tRNA gene, changing its anticodon to 5'UCA. B. A transition mutation in a tRNA gene, changing its anticodon to 5'CCA. C. A mutation in the amino-acyl tRNA synthetase that normally joins Trp to a tRNA. D. A transversion mutation in the same gene, upstream of the first mutation. E. A transversion mutation in the same gene, downstream of the first mutation.

A. A transition mutation in a tRNA gene, changing its anticodon to 5'UCA.

A31. Which of the following types of mutation, all within a long open reading frame, is least likely to cause a loss-of-function of the affected protein? A. A transition mutation in the third position of a codon near the 5' end of the transcript. B. A transversion mutation in the second position of a codon near the 3' end of a transcript. C. An insertion mutation in the second position of a codon, near the 5' end of the transcript D. A transversion mutation in the third position of a codon near the 3' end of the transcript. E. A deletion of the base at the second position of a codon near the 3' end of the transcript.

A. A transition mutation in the third position of a codon near the 5' end of the transcript.

A5. Which of the following statements about ribosomes is FALSE? A. Different ribosomes are needed to synthesize different proteins. B. Ribosomal RNAs are produced by transcription from a genomic DNA template. C. Most of the RNA in a cell is ribosomal RNA. D. Ribosomes are about half RNA and half protein, by mass. E. The formation of peptide bonds is catalyzed by a ribosomal RNA.

A. Different ribosomes are needed to synthesize different proteins.

A19. In a hypothetical punctuated genetic code, in which codons are marked off by specific chemical structures on the message, which type of mutation could NOT occur? A. Frameshift mutations B. Missense mutations C. Mutations affecting only a single amino acid D. Nonsense mutations E. Silent mutations

A. Frameshift mutations

C32. Which of the following statements about introns is FALSE? A. Introns are removed from the genomic DNA by a spliceosome before gene expression. B. Introns are bounded by short consensus sequences at their 5' and 3' ends. C. When compared across species, introns are more variable in sequence and length than coding exons. D. Introns probably originated as insertions of mobile genetic elements (transposons). E. Each intron is fully transcribed before being removed from the transcript. 7

A. Introns are removed from the genomic DNA by a spliceosome before gene expression.

M47. Which of the following statements about mismatch repair is FALSE? A. Mismatch repair of heteroduplex DNA formed by meiotic recombination is biased so that a mutant allele is always converted to the wild-type sequence. B. In post-replication mismatch repair in bacterial cells, methylated DNA indicates the strand that was used as the template for DNA replication. C. In post-replication mismatch repair in eukaryotic cells, misincorporation of ribonucleotides indicates the newly replicated DNA strand. D. A DNA base pair mismatch occurs when DNA polymerase adds a guanine nucleotide opposite a nucleotide containing adenine. E. Failure to correct a mismatch introduced during DNA replication can result in mutations.

A. Mismatch repair of heteroduplex DNA formed by meiotic recombination is biased so that a mutant allele is always converted to the wild-type sequence.

M44. Which of the following statements is FALSE? A. Mismatches that occur during DNA replication are repaired by replacing the nucleotide on the template strand with a nucleotide that is complementary to the mismatched base on the newly replicated strand of DNA. B. In bacteria, methylation is used to discriminate the template strand from newly replicated DNA during post-replication mismatch repair. C. In eukaryotic cells, misincorporation of ribonucleotides during replication is used to discriminate the template strand from newly replicated DNA during post-replication mismatch repair. D. A "G-T" mismatch in the heteroduplex DNA resulting from meiotic recombination is repaired randomly to either G-C or A-T. E. Post-replication mismatch repair increases the accuracy of DNA replication.

A. Mismatches that occur during DNA replication are repaired by replacing the nucleotide on the template strand with a nucleotide that is complementary to the mismatched base on the newly replicated strand of DNA.

A13. The genetic code is redundant or degenerate. This means that: A. More than one codon can encode the same amino acid B. More than one amino acid can be encoded by the same codon C. Almost all organisms use the same genetic code D. More than one nucleotide encodes an amino acid E. Codons overlap so that one nucleotide can form part of more than one codon.

A. More than one codon can encode the same amino acid

B5. Which of the following statements about translation and the genetic code is TRUE? A. Stop codons are codons for which there is no tRNA with a complementary anticodon. B. Code specificity is due to a chemical affinity between each amino acid and the codon that encodes it. C. The genetic code is redundant, meaning that each codon may encode more than one amino acid. D. The information to specify the primary structure of a polypeptide is provided by the ribosome that makes it. E. A mutant tRNA with an anticodon that is complementary to a stop codon will stop translation.

A. Stop codons are codons for which there is no tRNA with a complementary anticodon.

C1. The figure shows Remdesivir, a drug used for Covid-19 therapy. How is it most likely to act? A. Stopping viral genome replication by a viral RNA polymerase, after incorporation into RNA. B. Binding to the cell-surface ACE2 protein, preventing the virus from attaching to cells. C. Stopping viral genome replication by a viral DNA polymerase, after incorporation into DNA. D. Binding to the viral spike protein, preventing it from attaching to cell surface ACE2 proteins. E. Inhibiting a viral protease that cuts the viral polyprotein into separate functional proteins.

A. Stopping viral genome replication by a viral RNA polymerase, after incorporation into RNA.

P24. A scientist independently identified two different yeast mutants which are auxotrophic for adenine which she called ade-1 and ade-2. She constructs a diploid by crossing the ade-1 strain with the ade-2 strain and observes that the diploid fails to grow on medium lacking adenine. After putting the diploid through meiosis, she plates the cells onto rich medium that contains adenine and minimal medium lacking adenine. She obtains the following results: # colonies on rich medium + adenine: 100,000 # colonies on medium lacking adenine: 100 Which of the following statements is TRUE? A. The ade-1 mutation affects a different nucleotide basepair within the same gene as the ade-2 mutation. B. The ade-1 mutation affects the same nucleotide basepair within the same gene as the ade-2 mutation. C. The ade-1 mutation complements the ade-2 mutation. D. The ade-1 mutation is in a different gene from the ade-2 mutation.

A. The ade-1 mutation affects a different nucleotide basepair within the same gene as the ade-2 mutation.

M28. Which of the following is FALSE about eukaryotic DNA replication? A. The direction of post-replicative mismatch correction is random. B. Replication is initiated at multiple sites on a chromosome and proceeds bi-directionally. C. In most cells, replication of chromosomes is limited to once per cell cycle. D. Replication of the lagging strand occurs in the 5'3' direction. E. Replication of the leading strand occurs in the 5'3' direction.

A. The direction of post-replicative mismatch correction is random.

C29. The electron micrograph and sketch show the result from a 1977 experiment by Phillip Sharp, in which the genomic DNA of adenovirus, a eukaryotic virus, was separated into single strands and hybridized to polyadenylated mRNA from adenovirus-infected cells. It shows a structure in which a double-stranded DNA-RNA hybrid segment (heavy line) is interrupted by three loops of single-stranded DNA (numbered). Each single-stranded loop is: A. The template strand of an intron. B. The sense strand of an intron. C. The template strand of an exon. D. The sense strand of an exon. E. A poly-A tail.

A. The template strand of an intron.

P8. A yeast strain auxotrophic for uracil is transformed with four different plasmids each carrying a wild type copy of a different gene encoding an enzyme involved in the synthesis of uracil (designated URA1, 3, 5 and 6). The transformed cells were plated onto medium lacking uracil and the following results were obtained: Plasmid Added Growth on -ura medium none no colonies URA1 no colonies URA3 no colonies URA5 200 colonies URA6 no colonies What can we conclude from this experiment? A. This yeast strain is mutant in the ura5 gene. B. This yeast strain in mutant in the ura3 gene. C. The URA5 gene acts first in the pathway to synthesize uracil. D. The URA5 gene acts last in the pathway to synthesize uracil. E. This yeast strain is mutant in more than one gene involved in synthesizing uracil.

A. This yeast strain is mutant in the ura5 gene.

B20. In a thalassemia (blood disease) prevalent in Southeast Asia, the hemoglobin A polypeptide chain is longer than normal, but the mRNA length is unchanged. The normal HbA polypeptide ends with the sequence: - Lys - Tyr - Arg -(C terminus). The mutant version adds 30 amino acids, ending with: - Lys - Tyr - Arg - Gln - (29 more amino acids) -(C terminus). The causative mutation in the HbA gene is most likely: A. a transition mutation. B. a transversion mutation. C. a nonsense mutation. D. a frameshift mutation. E. an insertion mutation

A. a transition mutation.

O18. The following genes are on the X-chromosome of Drosophila in the order given: y-cv-ct-sn-m where the mutant alleles are all recessive. y results in yellow body color, cv results in crossveinless wings, ct results in cut wings, sn results in singed bristles, and m results in miniature wings. Wild-type males were irradiated with X-rays and mated with females homozygous for y cv ct sn and m. Most of the female progeny from this cross were phenotypically wild-type but one female exhibited crossveinless cut wings and singed bristles (cv ct sn). What is the nature of the X-ray induced mutation that resulted in the exceptional female? A. deletion B. inversion C. translocation D. dicentric E. acentric

A. deletion

P23. The following E. coli mutants were tested for growth on four known precursors of thymine, A-D. "+" indicates growth, "-" indicates no growth. Based on these data, the mutants define genes that act in a linear pathway for the biosynthesis of thymine in which order? Mutant Precursor/product A B C D thymine 9 + - + - + 10 - - + - + 14 + + + - + 21 - - - - + A. precursor-14-9-10-21-thymine B. precursor-21-10-9-14-thymine C. precursor-9-10-21-14-thymine D. precursor-14-10-9-21-thymine E. precursor-21-9-14-10-thymine

A. precursor-14-9-10-21-thymine

O30. Nitrous acid is a mutagen that deaminates adenine to form hypoxanthine, a modified base that pairs with cytosine. The MOST LIKELY type of mutation caused by nitrous acid is a: A. transition mutation B. transversion mutation C. inversion D. translocation E. duplication

A. transition mutation

O14. In fluctuation test experiments, bacterial mutation to phage resistance is examined by growing separate bacterial cultures for several generations, then plating the bacteria together with the phage. The number of resistant colonies per culture fluctuates (varies) between cultures: most yield a few resistant colonies, but some cultures had many. These results imply that mutations to phage resistance.... A. - are induced by phage infection, but at different rates in different cultures. B. - happen independently of phage infection, but at different times in different cultures. C. - are induced by phage infection, but at different times in different cultures. D. - happen independently of phage infection in most cultures, but are induced by phage infection in some cultures.

B. - happen independently of phage infection, but at different times in different cultures.

P1. A yeast geneticist isolates six independent mutants (1-6) which fail to grow on medium lacking histidine. She crosses each mutant to itself as well as to the other five mutants and assays the ability of the resulting diploids to grow on -his medium. She obtains the following data (+ = diploid is prototrophic for histidine, - = diploid is auxotrophic for histidine). a mating type 1 2 3 4 5 6 1 - + - + - + 2 + - + + + + mating type 3 - + - + - + 4 + + + - + - 5 - + - + - + 6 + + + - + - From these data, the geneticist places the different mutants into complementation groups. How many mutants are present in the complementation group which contains mutant #4? (Include mutant #4 as one of the alleles). A. 1 B. 2 C. 3 D. 4 E. 5

B. 2

O27. Cycloheximide is a chemical that inhibits protein synthesis, thereby killing yeast cells. Mutations in CYH2, a gene that encodes a ribosomal protein that is unable to bind cycloheximide, make cells resistant to the chemical. 1 ml of culture contains 108 yeast cells. 0.1 ml of this culture is plated onto medium containing cycloheximide and 2 colonies grow up. What is the frequency of cycloheximide resistant cells in the culture? A. 2 X 10-8 B. 2 X 10-7 C. 1 X 10-8 D. 4 X 10-7 E. 4 X 10-8

B. 2 X 10-7

P36. In yeast, mating requires the activation of two protein kinases encoded by the STE11 and STE7 genes. Ste11 phosphorylation of Ste7 activates Ste7 and the activated Ste7 then phosphorylates downstream proteins. Deletion of either STE11 (ste11∆) or STE7 (ste7∆) makes cells sterile, that is unable to mate. Constitutively active mutant versions of STE11 (STE11-c) and STE7 (STE7-c) exist (that is, they are active under all conditions) and are able to mate. Based on this information which of the predictions is MOST LIKELY to be TRUE? A. A haploid containing ste11∆ ste7∆ will be able to mate. B. A haploid containing ste11∆ STE7-c will be able to mate. C. A haploid containing STE11-c ste7∆ will be able to mate. D. A haploid containing STE11-c STE7-c will be unable to mate. E. A haploid containing STE11 STE7 will be unable to mate.

B. A haploid containing ste11∆ STE7-c will be able to mate.

C33. Which of the following statements about transcription in eukaryotic cells is FALSE: A. Different RNA polymerases transcribe protein-coding and non protein-coding genes. B. All of the DNA in a eukaryotic cell is transcribed. C. A long stretch of a repeated nucleotide is added to the 3' end of some transcripts D. Non-coding segments of RNA are removed from some transcripts before translation. E. Protein-coding RNAs are transcribed in the nucleus and exported to the cytoplasm for translation.

B. All of the DNA in a eukaryotic cell is transcribed.

P42. A biochemical pathway for the synthesis of red flower color proceeds as follows: Enzyme A Enzyme B Enzyme H Yellow intermediate -----> Orange intermediate ----> Blue intermediate ----> Red pigment Enzymes A, B and H are encoded by genes A, B and H, respectively. This plant is diploid and all three genes are on different chromosomes. Two plants with the following genotypes are crossed: AA bb Hh X aa BB hh Based on this information, which of the following statements is FALSE? A. The parental plant with the genotype AA bb HH has orange flowers. B. All of the progeny plants from this cross will exhibit red flowers due to complementation. C. One half of the progeny plants will exhibit blue flowers. D. The parental plant with the genotype aa BB hh has yellow flowers. E. The three genes independently assort during meiosis.

B. All of the progeny plants from this cross will exhibit red flowers due to complementation.

O12. Which of the following statements is NOT TRUE? A. Bacterial cells resistant to antibiotics are pre-existing in a culture and are selected upon exposure to the antibiotics. B. Bacterial cells exposed to antibiotics mutate in response to the antibiotics to become resistant. C. Mutants in genes that inactivate DNA repair pathways such as base excision and nucleotide excision repair exhibit higher frequencies of mutations. D. Intercalating agents such as ethidium bromide or acridine orange frequently result in the insertion or deletion of one or more basepairs.

B. Bacterial cells exposed to antibiotics mutate in response to the antibiotics to become resistant.

P18. Given the pedigree for two families that have people affected by deafness (indicated by shaded in circles or squares), which of the following statements is NOT TRUE? A. Individual II-2 is most likely homozygous for a recessive mutation that causes deafness. B. Because two affected people had unaffected children, individuals II-7 and II-8 must be heterozygous for an autosomal dominant mutation that causes deafness. C. This pedigree demonstrates that mutations in two different genes can cause deafness. D. The probability that individual II-9 is heterozygous for a mutation causing deafness is 2/3. E. The fact that the II-7 male had unaffected daughters rules out that this trait is due to an X-linked dominant mutation.

B. Because two affected people had unaffected children, individuals II-7 and II-8 must be heterozygous for an autosomal dominant mutation that causes deafness.

M46. Which of the properties of DNA polymerase can best explain why semi-conservative DNA replication requires both leading and lagging strand DNA synthesis? A. Proofreading activity of DNA polymerase B. DNA polymerase can only add nucleotides in the 5' to 3' direction. C. DNA polymerase requires a template. D. DNA polymerase requires a primer. E. DNA polymerase requires nucleotides containing adenine, guanine, cytosine and thymine.

B. DNA polymerase can only add nucleotides in the 5' to 3' direction.

A12. In experiments that showed the genetic code to be a triplet, reading frame code, Francis Crick and co-workers classified mutations in the rII gene of bacteriophage into two groups. Some combinations of two mutations from different groups, when recombined into the same gene, could suppress each other, but no two mutations from the same group could suppress each other. These two groups represented: A. Transition and transversion mutations. B. Insertion and deletion mutations. C. Loss-of-function and gain-of-function mutations. D. Mutations in codons and in anticodons. E. Missense and nonsense mutations. 2

B. Insertion and deletion mutations.

M9. Replicating structures in DNA can be observed in the electron microscope. Regions being replicated appear as bubbles. A part of a replicating eukaryotic chromosome is shown below. There are three different origins of replication numbered 1, 2 or 3. Assuming that all replication forks move at the same speed, which origin of replication was activated last? ==000000==00==0000== (1,2,3 respectively) A. Origin 1 B. Origin 2 C. Origin 3 D. Cannot be determined based on the information given

B. Origin 2

O37. Muller irradiated male fruit flies with X-rays and crossed them to females heterozygous for a balancer X chromosome containing the dominant Bar eyes mutation. He took the F1 Bar eyed female progeny from this cross and mated individual females with wild-type, non-irradiated males. When he scored the F2 progeny from each cross, he found that in most crosses, there were equal numbers of normal eye: Bar eyed males. There were some crosses, however, in which only Bar eye males were observed. What is the reason that these crosses did not exhibit any normal eyed males? A. Bar eyes is epistatic to Bar+ and masked the normal eye phenotype. B. Some of the F1 females contained deletions of essential genes on the chromosome inherited from the irradiated male. C. Crossovers on the X chromosomes of some of the F1 females resulted in homozygosis of the Bar gene. D. Meiosis I non-disjunction in the female. E. The Bar mutant is pleiotropic.

B. Some of the F1 females contained deletions of essential genes on the chromosome inherited from the irradiated male.

B19. A gene encoding a tRNA with the anticodon 5'-GUA-3' is mutated so that the anticodon is now 5'-CUA-3'. In a heterozygous cell producing both the normal and mutant versions of this tRNA, which of the following consequences is most likely, compared to a cell producing only the normal tRNA? A. Some proteins will be shortened. B. Some proteins will be lengthened. C. Some proteins will have leucines inserted instead of valines. D. Some proteins will have valines inserted instead of leucines. E. There will be no change in any of the proteins produced.

B. Some proteins will be lengthened.

O11. In cells treated with an oxidation agent, guanines may be converted to 8-oxodG, a modified base that pairs with adenine. Assuming that this happens, what type of mutation is the MOST LIKELY outcome if the cells are allowed to undergo two rounds of cell division? A. a transition mutation B. a transversion mutation C. an insertion mutation D. a deletion mutation E. an inversion mutation

B. a transversion mutation

M19. The bacterial repair system that corrects mismatched bases after DNA replication is able to discriminate between the old and newly made DNA strands because: A. older DNA is more likely to contain errors. B. older DNA contains methyl groups at specific sequences. C. new DNA contains methyl groups at specific sequences. D. DNA polymerase is covalently attached to the new strand.

B. older DNA contains methyl groups at specific sequences.

C6. In the sense strand of a bacterial gene, 35% of the bases are adenine. From this you can be sure that among the bases in the transcript of this gene there will be: A) 35% guanine. B) 35% cytosine. C) 35% adenine. D) 35% thymine. E) 35% uracil.

C) 35% adenine.

B16. A peptide translated from an mRNA in a cell-free system normally has a sequence beginning: Met - Ala - Arg - Cys -. If an amino-acyl-tRNA with the anticodon 5'-GCA-3' is chemically altered, so that the amino acid it carries is now Ala, and is added to the system, you would expect to find altered peptides that begin: A) Met - Cys - Arg - Ala - B) Met - Ala - Ala - Ala - C) Met - Ala - Arg - Ala - D) Met - Cys - Arg - Cys - E) Met - Ala - Ala - Cys -

C) Met - Ala - Arg - Ala -

B9. According to the "wobble" rules for codon-anticodon pairing, inosine (I) in the 5' position of an anticodon can pair with U, C or A at the complementary position in the codon. Which of the following amino acids is most likely to be translated using a tRNA with inosine in the 5' position of its anticodon? A) Met B) His C) Pro D) Tyr E) Trp

C) Pro

C40. The human dystrophin gene is 2.3 million basepairs long and includes 79 exons. The mRNA that is produced from this gene is 14,000 bases long. If transcription proceeds at 2400 bases per minute, about how long does it take to produce one complete dystrophin mRNA? A. 6 minutes B. 80 minutes C. 16 hours D. Cannot be estimated from this information because it depends primarily on the splicing rate.

C. 16 hours

O24. 1 ml of culture contains 108 yeast cells. 0.1 ml of this culture is plated onto medium containing canavanine and 20 colonies grow up. What is the frequency of canavanine-resistant cells on the culture? A. 2 X 10-7 B. 4 X 10-7 C. 2 X 10-6 D. 4 X 10-6 E. 1 X 10-6

C. 2 X 10-6

P2. A yeast geneticist isolates six independent mutants (1-6) that fail to grow on medium lacking lysine. She crosses each mutant to itself as well as to the other five mutants and assays the ability of the resulting diploids to grow on medium lacking lysine. She obtains the following data (+ = diploid is prototrophic for lysine - = diploid is auxotrophic for lysine). MATa mating type 1 2 3 4 5 6 1 - + - + - + 2 + - + + + + MAT mating type 3 - + - + - + 4 + + + - + - 5 - + - + - + 6 + + + - + - From these data, the geneticist places the different mutants into complementation groups. How many mutants are present in the complementation group that contains mutant #1? (Include mutant #1 as one of the alleles). A. 1 B. 2 C. 3 D. 4 E. 5

C. 3

P4. The yeast Saccharomyces cerevisiae was mutagenized and five mutant strains (a, b, c, d and e) were obtained that are auxotrophic for the amino acid lysine. Complementation tests were performed by mating each mutant to itself as well as to each of the four other mutants. The resulting diploids were tested for their ability to grow on medium lacking lysine. Growth is indicated by + and no growth is indiciated by - in the following table. How many different genes did this experiment isolate that are involved in lysine biosynthesis? a b c d e a - + + + - b + - + + + c + + - - + d + + - - + e - + + + - A. 1 B. 2 C. 3 D. 4 E. 5

C. 3

P5. In Neurospora mutants A, B, C, D, E, F and G all have the same phenotype: auxotrophy for histidine. In pairwise combinations in complementation tests, the following results were produced, where + = complementation and - = no complementation. How many different complementation groups are defined by this analysis? A B C D E F G G + - + + + + - F - + + - + - E + + - + - D - + + - C + + - B + - A - A. 1 B. 2 C. 3 D. 4 E. 5

C. 3

P21. A biochemical pathway for the synthesis of red flower color is found to proceed as follows: Enzyme A is encoded by gene A, enzyme Q is encoded by gene Q, enzyme H is encoded by gene H. This particular plant is diploid and all three genes are on different chromosomes. Two plants with the following genotypes are crossed: Aa Qq hh X aa Qq Hh What is the probability of a plant from this cross producing red flowers? A. 1/4 B. 7/8 C. 3/16 D. 1/2 E. 3/8

C. 3/16

A26. How many different amino-terminal tripeptide (three-amino-acid) sequences can theoretically be generated by translation in eukaryotes? A. 60 B. 380 C. 400 D. 8000 E. 41

C. 400

B11. Which of the following anticodons can be found on a tRNA attached to the amino acid aspartate (Asp)? A. 5' IUC -3' B. 5' CUI-3' C. 5' GUC-3' D. 5' CUG-3' E. 5' GUI-3'

C. 5' GUC-3'

M14. Assume that a replication fork starts at the left of the DNA sequence shown below and moves towards the right. What is the sequence of the DNA synthesized as the lagging strand? 5' GGTATTCATCT 3' 3' CCATAAGTAGA 5' A. 5' GGTATTCATCT 3' B. 5' TCTACTTATGG 3' C. 5'AGATGAATACC 3' D. 5'CCATAAGTAGT 3'

C. 5'AGATGAATACC 3'

P37. The R+ gene encodes an enzyme used to make red pigment in a particular type of flower. Red color requires 20 units of enzyme in a cell. If there is less than 11 units of enzyme/cell, the flowers are white. Between 11-19 units/cell the flowers are pink. One R+ allele generates 10 units of enzyme/cell, while one mutant r allele makes 5 units/cell. Which of the following statements about this plant is FALSE? A. A plant that is heterozygous for a deletion of the R+ gene (R+/r∆) is haploid insufficient for flower color. B. The R+ allele is incompletely dominant to the r allele. C. A diploid that is heterozygous r/r∆ will exhibit a pink phenotype due to pseudodominance. D. A cross between a purebreeding red flowered plant and a plant homozygous rr will result in F1 plants that are all pink. E. Plants that are homozygous rr will exhibit the same flower color phenotype as plants homozygous for the deletion (r∆/r∆).

C. A diploid that is heterozygous r/r∆ will exhibit a pink phenotype due to pseudodominance.

B2. Which of the following statements about tRNAs is FALSE? A. A given tRNA will always be joined to the same amino acid. B. Some amino acids can be joined to more than one species of tRNA. C. A mutant amino-acyl-tRNA with the anticodon 5'UCA3' will stop translation. D. Some tRNAs can base-pair with more than one codon. E. Wild type cells contain no tRNAs that base-pair with stop codons.

C. A mutant amino-acyl-tRNA with the anticodon 5'UCA3' will stop translation.

C37. Which of the following is not directly encoded in genomic DNA? A. Ribosomal RNAs (rRNA). B. Transfer RNAs (tRNA). C. A poly-A tail on a eukaryotic mRNA. D. Signal peptides. E. None of these are encoded in genomic DNA.

C. A poly-A tail on a eukaryotic mRNA.

O10. Which of the following statements about DNA repair is FALSE? A. Most damage to DNA is repaired. B. Mutations in DNA repair enzymes can predispose to cancer. C. DNA repair mechanisms can distinguish between DNA sequences that encode functional versus non-functional proteins. D. Some defects in DNA repair are associated with premature aging. E. Some DNA repair mechanisms can distinguish between newer and older DNA strands.

C. DNA repair mechanisms can distinguish between DNA sequences that encode functional versus non-functional proteins.

M5. Which statement is NOT true with regard to lagging strand DNA synthesis? A. DNA synthesis occurs in the 5'-3' direction. B. Deoxyribonucleotides are added to an RNA primer. C. DNA synthesis is continuous. D. Okazaki fragments are formed. E. DNA polymerase adds deoxyribonucleotides that are complementary to the template strand of DNA.

C. DNA synthesis is continuous.

A30. Which one of the following alterations, all in the protein-coding part of a transcript, is most likely to eliminate the activity of the protein it normally encodes? A. G to C in the 3rd position of a codon near the 5' end of the transcript B. Insertion of an A into the 2nd position of a codon near the 3' end of the transcript C. Deletion of a U in the 3rd position of a codon near the 5' end of the transcript D. C to U transition in the 1st position of a codon near the 3' end of the transcript

C. Deletion of a U in the 3rd position of a codon near the 5' end of the transcript

A21. Two single-base substitutions in the trpA gene of E.coli both affected amino acid number 234. In one mutant, the amino acid at this position was now Cys, in the other mutant it was Asp. What was the most likely amino acid encoded at position 234 in the wild type protein? A. Val B. Trp C. Gly D. Arg E. Pro

C. Gly

M11. Which of the following statements is FALSE about DNA replication? A. DNA replication requires a template strand and an oligonucleotide primer (either RNA or DNA). B. Deoxyribonucleotides are added to the 3' end of an oligonucleotide primer. C. In a bidirectional replication fork, the direction of DNA synthesis on the lagging strand is towards the replication fork. D. The lagging strand is synthesized discontinuously to produce Okazaki fragments. E. In most cells, DNA replication is regulated such that it occurs at one stage of the cell cycle.

C. In a bidirectional replication fork, the direction of DNA synthesis on the lagging strand is towards the replication fork.

P41. A yeast geneticist isolated five mutants that are auxotrophic for arginine. She introduced the mutations, indicated 1, 2, 3, 4 and 5 into haploid MATa and MAT strains, crossed them to make diploids, patched out the diploids on agar plate containing rich medium and then transferred the cells onto medium lacking arginine and put the cells in the incubator to grow. The next day she scored which patches could grow or not and returned the plate to the incubator. Five days later she scored the patches again and discovered a few colonies growing for the 1 X 4 diploid. Black indicates cell growth, while white indicates no growth. Based on these data, which of the following statements is FALSE? A. There are at least three genes that are required for making arginine. B. Mutant # 1 and mutant # 4 are alleles of the same gene. C. Mutant # 4 is mutated at the same nucleotide position as mutant #1. D. Mutant # 2 complements mutant #4. E. Mutant # 2 is mutated at the same nucleotide position as mutant # 3.

C. Mutant # 4 is mutated at the same nucleotide position as mutant #1.

M10. Which of the following factors does not play a role in the high fidelity (low number of errors) observed for replicated molecules of DNA? A. complementary base-pairing B. 3'-5' proofreading activity of DNA polymerase C. Okazaki fragments D. Mismatch repair

C. Okazaki fragments

A17. Protein sequencing showed (i) that any amino acid can be followed by any other, and (ii) that a mutation can change a single amino acid in a protein. Which hypothetical genetic codes can be excluded by either of these observations? A. Punctuated codes, in which the codons are indicated by chemical marks on the DNA. B. Reading-frame codes, in which a specific starting point sets the frame in which the code will be read. C. Overlapping codes in which neighboring codons share two of their three bases. D. Codes in which each codon includes more than one base. E. Codes in which each codon is a single base (and there are 20 different bases).

C. Overlapping codes in which neighboring codons share two of their three bases.

O23. A scientist performed the Ames test on four different substances to see whether anything of them are mutagenic. She incubated the substances with his- bacteria alone, or with his- bacteria to which mammalian liver enzymes were added. Which of the following conclusions is NOT supported by the data she obtained below? Number of His+ colonies his- bacteria his- bacteria plus liver enzymes No substance 0 0 Substance 1 120 0 Substance 2 0 0 Substance 3 95 102 Substance 4 0 125 A. Substance 2 is unlikely to be mutagenic in mammals. B. Substance 3 is likely to be mutagenic in mammals and should be studied further. C. Substances 1 and 4 are likely to be mutagenic in mammals and should be studied further. D. Mammalian liver enzymes convert Substance 4 to a compound that is likely to be mutagenic in mammals. E. Mammalian liver enzymes have little to no effect on the mutagenicity of Substance 3.

C. Substances 1 and 4 are likely to be mutagenic in mammals and should be studied further.

M52. Which of the following statements is FALSE? A. When a mismatched basepair is created during DNA replication, it is the nucleotide on the newly synthesized strand that is replaced by the mismatch repair machinery. B. DNA synthesis on the lagging strand occurs in the 5' to 3' direction. C. The direction of DNA synthesis on the leading strand is away from the replication fork. D. When a mismatched base pair is created during meiotic recombination, it is random which nucleotide is replaced by the mismatch repair machinery. E. In most eukaryotic cells, DNA replication is regulated so that it occurs only once in each cell cycle.

C. The direction of DNA synthesis on the leading strand is away from the replication fork.

O19. A plant that is homozygous dominant for a particular gene, DD, is treated with X-rays and fertilized with pollen from a homozygous recessive plant, dd. Out of 500 offspring, two show the recessive phenotype. Which of the following events is most likely to give this result? A. X-rays induced reciprocal translocations in two of the gametes from the DD parental plant. B. X-rays induced pericentric inversions in two of the gametes from the DD parental plant. C. X-rays induced deletions in two of the gametes from the DD parental plant. D. X-rays induced duplications in two of the gametes from the DD parental plant. E. X-rays induced paracentric inversions in two of the gametes from the DD parental plant.

C. X-rays induced deletions in two of the gametes from the DD parental plant.

P30. In the biosynthetic pathway shown below, A, B and C represent intermediates in the biosynthesis of Final Product. mut1, mut2, mut3 and mut4 represent independently isolated mutants that fail to make Final Product. All of the mutants are recessive. Based on this pathway, which of the following statements is NOT TRUE? A. Final product can be obtained from a mut1 strain by adding back intermediate C. B. A mut3 mut1 double mutant will accumulate intermediate A. C. mut2 and mut4 must be mutations in the same gene. D. A diploid created by mating a mut3 haploid with a mut1 haploid will make Final Product. E. A mut4 haploid will accumulate intermediate C.

C. mut2 and mut4 must be mutations in the same gene.

O34. In fruit flies, w+ is required for red eyes. Part of the sequence of the w+ gene is 5'...ATTATT...3' 3'...TAATAA...5'. A recessive X-linked w allele that results in white eyes has the sequence 5'...ATTCTT...3' 3'... TAAGAA...5' To get red-eyed revertants, white-eyed males were mutagenized and crossed to white-eyed females to look for red-eyed progeny. Which of the mutagens listed in the table below is MOST LIKELY to revert the w allele to w+? Mutagen Mode of action X-rays Makes double strand breaks UV light makes thymine dimers oxidation converts guanine to 8-oxo-guanine which base pairs with adenine alkylation converts guanine to O-6-ethylguanine which base pairs with thymine A. X-rays B. UV light C. oxidation D. alkylation

C. oxidation

O38. Which of the following mutagens is most likely to lead to a transversion mutation? A. X-rays B. acridine orange (an intercalator) C. oxidation (changes guanine to 8-oxodG which base-pairs with adenine) D. 5-bromouracil (when incorporated in DNA, can change from base-pairing with adenine to pairing with guanine) E. nitrous acid (changes cytosine in DNA to uracil)

C. oxidation (changes guanine to 8-oxodG which base-pairs with adenine)

P13. A biochemical pathway for the synthesis of flower color is found to proceed as follows: Enzyme A Enzyme Q Enzyme H yellow orange blue intermediate intermediate intermediate red pigment Enzyme A is encoded by gene A, enzyme Q is encoded by gene Q, enzyme H is encoded by gene H. This particular plant is diploid. What color are the flowers of a plant with the genotype aaQqhh? A. red B. blue C. yellow D. orange E. multicolored

C. yellow

A9. Which of the following statements about the genetic code is FALSE? A) More than one codon can encode the same amino acid. B) Adjacent codons do not overlap, so any amino acid may be followed by any other in polypeptide sequences. C) The code is read in codons of three bases. D) The frame in which the code is read is set by gaps in the sequence of bases. E) The universality of the genetic code is evidence for a common origin of all organisms now alive.

D) The frame in which the code is read is set by gaps in the sequence of bases.

C24. Binding of a regulatory protein to a specific DNA site depends on interactions between the protein and: A) the 3'OH group on the ribose ring. B) the 5' carbon atoms in the backbone. C) negatively charged phosphate groups in the backbone. D) the base pairs at that site. E) DNA ligases that join the protein to the DNA.

D) the base pairs at that site.

C17. A consensus promoter sequence for bacterial genes is: A. ...an identical DNA sequence found just upstream of every bacterial gene. B. ...synthesized in the cell by RNA polymerase. C. ...the site recognized by rho protein together with RNA polymerase. D. ...composed of the bases occurring most often at conserved positions upstream from their start sites. E. ...the 35 base pairs immediately preceding the start codon. 6

D. ...composed of the bases occurring most often at conserved positions upstream from their start sites.

C31. Introns were discovered in R-loop hybridization experiments, which revealed sequences that are... A. ...present in polypeptides but absent from genomic DNA. B. ...present in mRNA but absent from genomic DNA. C. ...present in both polypeptides and genomic DNA but absent from mRNA. D. ...present in genomic DNA but absent from mRNA. E. ...present in polypeptides but absent from mRNA.

D. ...present in genomic DNA but absent from mRNA.

H4. The restriction enzyme Not1 recognizes and cuts the sequence 5'GC|GGCCGC3' in double-stranded DNA. The enzyme HaeIII recognizes and cuts the sequence 5'GG|CC3' (in each case "|" represents the cut site, and only one of the two strands is shown). In a random DNA sequence with all four bases present at equal frequency, what fraction of HaeIII restriction sites can also be cut by Not I? A. 1 (all of them) B. 1/8 C. 1/16 D. 1/256 E. 0 (none of them)

D. 1/256

O3. Assume that a wild-type sequence is 5' AGCCTAC 3'. Indicate the sequence in which a transversion mutation has occurred. A. 5' AGTCTAC 3' B. 5' AGCCGCCGCCGCCTAC 3' C. 5' AGCCCAC 3' D. 5' ATCCTAC 3' E. 5' AGCCTGC 3'

D. 5' ATCCTAC 3'

B12. Only one of the following tRNA anticodons can be used in cells that use the standard genetic code. Which? A. 5' IUU3' B. 5' UCA3' C. 5' CUA3' D. 5' IAC3' E. 5' IUA3'

D. 5' IAC3'

C41. Near their 3'ends, eukaryotic mRNAs have the polyadenylation signal 5'AAUAAA3', which is followed after about 30 bases by a poly-A tail. Which of the following sequences is most likely to occur in the template strand of a eukaryotic gene, at the end of the transcribed region? A. 5'-TTTTTTTTTT....30 bases....TTATTT -3' B. 5'-AATAAA....30 bases...AAAAAAAAA-3' C. 5'-AAAAAAAAA...30 bases....AAATAA-3' D. 5'-30 bases....TTTATT....-3' E. 5'-AATAAA....30 bases-3'

D. 5'-30 bases....TTTATT....-3'

B10. "Wobble" pairing allows inosine (I) in the 5' position of the anticodon to pair with the bases U, C or A in the codon. For which amino acid can there NOT exist a corresponding tRNA with inosine (I) in the 5' position of its anticodon? A. Ile B. Val C. Ser D. Cys E. Gly

D. Cys

M15. Okazaki fragments arise because: A. DNA polymerase needs a primer to initiate DNA replication. B. DNA replication can occur bi-directionally from a single origin of replication. C. mistakes in DNA replication are corrected by a mismatch repair system. D. DNA polymerase can only add nucleotides in a 5'-3' direction. E. of the need to add bases complementary to those of the template strand.

D. DNA polymerase can only add nucleotides in a 5'-3' direction.

M6. Which of the following statements best explains why DNA synthesis on the lagging strand is discontinuous? A. DNA polymerase can only add nucleotides in the 3' to 5' direction. B. Okazaki fragments must be covalently linked by DNA ligase. C. DNA synthesis can be bidirectional. D. DNA polymerase can only add nucleotides in the 5' to 3' direction. E. Mismatch repair can occur either in mitotically growing cells or meiotic cells.

D. DNA polymerase can only add nucleotides in the 5' to 3' direction.

M54. Which of the following statements is FALSE? A. During DNA replication, the lagging strand is synthesized in pieces called Okazaki fragments which are then ligated together. B. DNA polymerase adds nucleotides to the 3' end of an existing single strand of DNA or RNA. C. PCR uses a thermophilic DNA polymerase which remains active at the high temperature necessary to denature duplex DNA. D. DNA replication is conservative, such that both strands of the resulting sister chromatid are composed of newly synthesized DNA. E. Something that makes DNA replication less accurate will lead to an increase in mutations.

D. DNA replication is conservative, such that both strands of the resulting sister chromatid are composed of newly synthesized DNA.

M37. Which of the following statements is NOT TRUE about post replication mismatch repair? A. In eukaryotic cells, the newly replicated leading strand is detected by nicks introduced as the result of removing mis-incorporated ribonucleotides. B. In bacterial cells, the newly replicated DNA is unmethylated. C. In bacterial cells, post-replication mismatch repair is non-random so that the nucleotide on the newly synthesized strand is replaced. D. In eukaryotic cells, post-replication mismatch repair is non-random so that the nucleotide on the template strand is replaced. E. Mutants in genes required for post-replication mismatch repair exhibit an increased frequency of mutations.

D. In eukaryotic cells, post-replication mismatch repair is non-random so that the nucleotide on the template strand is replaced.

B4. In protein synthesis, which of the following does NOT contribute to correct translation of a specific codon? A. Recognition of a specific amino acid by an aminoacyl-tRNA synthetase. B. Recognition of a specific tRNA by an aminoacyl-tRNA synthetase. C. Specific base-pairing between the codon and an anticodon in a tRNA. D. Rejection of incorrect amino acid-tRNA combinations by the ribosome.

D. Rejection of incorrect amino acid-tRNA combinations by the ribosome.

M35. Which of the following statements about mismatch repair is FALSE? A. Mismatch repair occurs after DNA replication to correct errors that were introduced by DNA polymerase. B. Mismatch repair occurs during meiosis to correct base pair mismatches in heteroduplex DNA that have arisen as a result of recombination. C. The direction of mismatch repair in meiosis is random. D. The direction of mismatch repair occurring after DNA replication is random. E. Defects in genes required for mismatch repair result in an increased frequency of mutations.

D. The direction of mismatch repair occurring after DNA replication is random.

C19. A bacterial gene is mutated to delete the two base pairs at the -1 and -2 positions relative to the transcription start site. What is the most likely effect (if any), of this mutation on the transcript and the encoded polypeptide? A. No transcription and therefore no translation will occur. B. Transcription will start at a different position, the transcript will not be translated. C. The transcript and the polypeptide will be unchanged. D. Transcription will start at a different position, the polypeptide will be unchanged. E. Transcription will start at a different position, the polypeptide will be frameshifted.

D. Transcription will start at a different position, the polypeptide will be unchanged.

M3. The elongation of the leading strand during DNA synthesis: A. progresses away from the replication fork. B. occurs in the 3' to 5' direction. C. produces Okazaki fragments. D. depends on the action of DNA polymerase. E. does not require a template strand.

D. depends on the action of DNA polymerase.

P11. A biochemical pathway for the synthesis of red flower color is found to proceed as follows: Enzyme A Enzyme Q Enzyme H yellow orange blue intermediate intermediate intermediate red pigment Enzyme A is encoded by gene A, enzyme Q is encoded by gene Q, enzyme H is encoded by gene H. This particular plant is diploid. What color are the flowers of a plant with the genotype Aa qq HH? A. red B. blue C. yellow D. orange E. multicolored

D. orange

A8. Which of the following statements about the genetic code is FALSE? A) A sequence of three bases encodes a single amino acid. B) Neighboring codons are independent, they do not overlap. C) The reading frame is determined only by the translation start position. D) The code is nearly universal: the same code "dictionary" is used by almost all cells. E) Some codons can encode more than one amino acid

E) Some codons can encode more than one amino acid

A6. Which of the following statements about translation is FALSE? A) Any ribosome can translate any RNA base sequence between a start codon and the next stop codon. B) The reading frame that is used depends on the position of the start codon. C) Special enzymes join specific amino acids to specific tRNA molecules for use in translation. D) Translation in a eukaryotic cell takes place on ribosomes in the cytoplasm. E) The information that specifies the sequence of amino acids in a protein comes from the ribosome.

E) The information that specifies the sequence of amino acids in a protein comes from the ribosome.

C12. One way that bacteria adjust their gene expression to changing conditions is by activating different sigma proteins. Genes regulated by different sigma proteins probably have different: A) translation initiator codons B) operator sequences C) core RNA polymerases D) ribosome binding sites E) promoter sequences

E) promoter sequences

A16. A probe sent to the Earth-size planet Kepler 186-f transmits evidence of some very simple, nucleic acid-based life forms. On-board protein sequencing shows that 25 different amino acids make up the polypeptides in these organisms, but among all proteins sequenced, each amino acid is preceded or followed by only a subset of the others. This constraint suggests that the Keplerian code... A. ...is punctuated. B. ....uses four-base codons. C. ....is non-redundant. D ...uses two-base codons. E. ...uses overlapping codons.

E. ...uses overlapping codons.

C7. In the template strand of a bacterial gene, 22% of the bases in the transcribed region are adenine. Among the bases in the transcript of this gene there will be: A. 28% adenine. B. 22% adenine. C. 28% uracil. D. 22% thymine. E. 22% uracil.

E. 22% uracil.

O25. 1 ml of culture contains 108 yeast cells. 0.1 ml of this culture is plated onto medium containing canavanine and 400 colonies grow up. What is the frequency of canavanine-resistant cells in the culture? A. 2 X 10-5 B. 2 X 10-6 C. 4 X 10-3 D. 4 X 10-6 E. 4 X 10-5

E. 4 X 10-5

A24. A wild type polypeptide begins: (N)-Met Glu Tyr Ala Trp... A mutant form of this polypeptide begins: (N)-Met Glu Thr Arg Gly... What is the nature of the mutation? A. A single-base deletion from the second codon of the wild-type gene. B. A base substitution in the third codon of the wild-type gene. C. A single-base insertion in the third codon of the wild-type gene. D. A base substitution in the second codon of the wild-type gene. E. A single-base deletion from the third codon of the wild-type gene.

E. A single-base deletion from the third codon of the wild-type gene.

A23. Which of the following amino acids is encoded by codons that cannot all be recognized by a single tRNA? A. Met B. His C. Phe D. Trp E. Gly

E. Gly

P45. A scientist isolated five different mutant yeast strains (a, b, c, d, e) that were auxotrophic for the amino acid tryptophan. To see how many different genes were present in her mutant collection, she did a complementation test by mating different pairs of mutants to make diploids, which she then tested for their ability to grow in the absence of tryptophan. This experiment placed the mutant alleles into three complementation groups: Group 1: a,c / Group 2: b,e / Group 3: d. The biosynthetic pathway for tryptophan is where P1, P2 and P3 are intermediates. To see where in the pathway the different mutants were blocked, she tested each mutant for growth on minimal medium containing a different intermediate or tryptophan. The data are shown on the table with "+" indicating growth and "-" indicating no growth. Based on this information, which of the following statements is FALSE? Trp- mutant Added nutrient P1 P2 P3 Tryptophan a - + + + b - - + + c - + + + d ¬- - + + e - - + + A. Mutants a and c are in the same gene. B. Mutants d and e are in the same gene, which is required to convert P2 to P3. C. Mutants b and d are in different genes. D. Mutant b is in a gene required to convert P2 to P3. E. Mutant a is in a gene required to convert P1 to P2.

E. Mutant a is in a gene required to convert P1 to P2.

M33. Which of the following statements is FALSE? A. The genetic material in a chromosome is contained in one long duplex of DNA. B. One complete strand of DNA in the duplex of each sister chromatid is newly synthesized. C. Duplicated sequences on the same chromosome may expand or contract by unequal crossing over. D. Tetraploid organisms can produce euploid gametes. E. Polyploid species are aneuploid.

E. Polyploid species are aneuploid.

P7. Two haploid yeast strains, each auxotrophic for leucine, are mated to make a diploid. The diploid is tested for growth on medium lacking leucine and the two mutants complement each other. Based on these results, which statement is TRUE? A. The diploid cells were not able to grow on the medium lacking leucine. B. If the diploid is put through meiosis, spores auxotrophic for leucine are not likely to be observed. C. The two mutants are caused by the same change in the DNA sequence of the same gene. D. The two mutants are caused by different changes in the DNA sequence of the same gene. E. The mutants are in two different genes.

E. The mutants are in two different genes.

O5. A deamination event converts the adenine indicated by the arrow in the DNA strand below to hypoxanthine, a base that behaves like guanine with respect to complementary base pairing. After ONE round of DNA replication, the mutation that will be observed is a: A. deletion B. a T to G transversion C. an A to C transversion D. an A to G transition E. a T to C transition

E. a T to C transition

P12. A biochemical pathway for the synthesis of flower color is found to proceed as follows: Enzyme A Enzyme Q Enzyme H yellow orange blue intermediate intermediate intermediate red pigment Enzyme A is encoded by gene A, enzyme Q is encoded by gene Q, enzyme H is encoded by gene H. This particular plant is diploid. What color are the flowers of a plant with the genotype aa Qq Hh? A. red B. blue C. multicolored D. orange E. yellow

E. yellow

C14. Which of the following statements about bacterial promoters is FALSE: A) They include an AT-rich sequence located 10bp upstream of the transcription start site. B) RNA polymerase and sigma protein bind to them to initiate transcription. C) They determine the sites at which transcription will begin. D) Several different genes can be transcribed from a single promoter. E) They are sites of sequence-specific interaction between DNA and proteins. F) They are located a fixed distance upstream from a transcription start site. G) They determine the reading frame for translation of the message. H) Different genes have similar, but not necessarily identical, promoter sequences. I) They are not themselves transcribed.

G) They determine the reading frame for translation of the message.