Final Exam (12/12/2013)

1. (2 pts.) The presence in a population of more than two alleles of a gene is called

A. Dihybrid
B. Pleiotropic
C. Multiple alleles
D. Mutant
E. Codominant

2. (2 pts.) The coupling relationship or phase (coupling or repulsion) of alleles in a doubly heterozygous parent is detectable as the

A. Largest number of map units between corresponding alleles
B. Lowest incidence of crossing over
C. Most frequent type of gametes
D. Highest frequency of recombination
E. Double recessive type

3. (2 pts) The enzyme activity of DNA polymerase that functions in proofreading is

A. 3' to 5' exonuclease
B. 5' to 3' exonuclease
C. DNA ligase
D. Glycosylase
E. Flap endonuclease

4. (2 pts) DNA ligase functions to

A. Catalyze the formation of hydrogen bonds between adjacent 5'-P and 3'-OH termini
B. Catalyze the formation of covalent bonds between adjacent 5'-P and 3'-OH termini
C. Facilitate base pairing between single-stranded molecules of DNA
D. Relax supercoiling of DNA
E. Add methyl groups to DNA

5. (2 pts) Which of the following statements about PCR is false?

A. PCR uses short oligonucleotide primers
B. PCR stands for polymerase chain reaction
C. PCR can be used to obtain large quantities of a particular DNA sequence
D. PCR does not require knowledge of the DNA sequences at the ends of the region to be amplified
E. PCR uses a DNA polymerase to synthesize DNA

6. (2 pts) Which of the following features is common to both DNA replication and transcription?

A. Deoxyribonucleotides are incorporated into the growing sequence
B. Nucleotides are added to the 5' end of the newly synthesized strand
C. A sugar-phosphate bond is formed between the 3' hydroxyl and the 5' phosphate
D. Both RNA and DNA polymerase require oligonucleotide priming
E. Both RNA and DNA polymerase initiate at promoter sequences

7. (2 pts) Which of the following statements is true regarding gene expression?

A. The first step is the association of mRNA with ribosomes
B. Involves proofreading of the mRNA
C. The 3' end of the mRNA corresponds to the carboxyl terminus of the protein
D. Polypeptides are synthesized by addition of amino acids to the amino terminus
E. Prokaryotic RNA usually undergoes nuclear processing

8. (2 pts) Which of the following statements is true?

A. tRNA has a 5' triphosphate
B. tRNAs are charged by binding to the ribosome
C. There are more distinct tRNAs than codons
D. Amino acids are bound to the 5' end of charged tRNAs
E. The anticodon of one tRNA can bind with several different codons

9. (2 pts) Which of the following statements about translation is correct?

A. In translation initiation, an aminoacyl tRNA binds the start codon on the mRNA before the large ribosomal subunit binds.
B. Translation termination occurs when an uncharged tRNA recognizes the stop codon in the A site.
C. Translation termination occurs when the ribosome reaches the 3’ end of the mRNA.
D. In bacterial translation initiation, sigma factor recruits the large ribosomal subunit.
E. Peptide bonds are synthesized between a polypeptide chain attached to the large ribosomal subunit and an aminoacyl tRNA in the E site.

10. (2 pts) Which of the following statements about mutations is correct?

A. In a frameshift mutation, deletion or addition of three bases causes a single amino acid to be gained or lost.
B. In a splice site mutation, multiple amino acids may be missing from the protein product.
C. In a missense mutation, the stop codon is changed to a codon for an amino acid.
D. In a silent mutation, transcription of the gene is silenced.
E. In a nonsense mutation, multiple amino acids are added to the carboxyl terminus compared to the wild-type protein sequence.

11. (2 pts) Which of the following statements about forensic analysis of short terminal repeats (STRs) is false?

A. The probability of two unrelated people sharing a DNA profile derived from the analysis of 14 standard markers is about one in a billion.
B. Short terminal repeats consist of a variable number of repeats of a short DNA sequence flanked by unique sequence.
C. Saliva from a cigarette butt, licked envelope, or chewing gum contains enough human DNA to conduct forensic analysis.
D. Primers that recognize the unique sequence flanking STRs are used to amplify DNA using PCR.
E. Amplified DNA is analyzed by mass spectrometry.

12. (2 pts) Mutations are

A. Heritable changes in genetic information
B. Usually beneficial to organisms
C. Caused by genetic recombination
D. Caused by errors in translation
E. Caused by errors in transcription

13. (2 pts) Which of the following statements is true regarding generalized transduction?

A. It requires direct contact between donor and recipient cells
B. Some phage particles may contain bacterial DNA only
C. Phage integrate at a specific site in the bacterial chromosome
D. Transfer involves DNA from a particular region of the bacterial chromosome
E. None of the above

14. (2 pts) In conjugation

A. Plasmids replicate prior to conjugative transfer
B. Conjugation pili are synthesized only by the recipient cell
C. Genetic exchange is bidirectional between conjoined cells
D. Chromosomal genes from the donor cell are transferred only when the F factor integrates into the donor chromosome
E. The fertility factor is transferred as a supercoiled double-stranded circular DNA

15. (2 pts) Which of the following statements is false regarding Hfr transfer?

A. F- cells undergo recombination with regions of donor DNA
B. Hfr transfer can be used to create a genetic map
C. The donor genotype becomes F-
D. The recipient remains F-
E. Large regions of the donor chromosome are transferred

16. (2 pts) A negatively regulated genetic system

A. Cannot encode enzymes involved in biosynthesis
B. Is more common in eukaryotes
C. Is completely off when repressor protein is present
D. Has a lower level of transcription when a regulatory protein interacts with the regulatory region of the gene or operon
E. All of the above

17. (2 pts) A positively regulated genetic system

A. Must override an inhibitor
B. Is more common in prokaryotes
C. Is derepressed by inducer
D. Has a higher level of transcription when a regulatory protein interacts with the regulatory region of the gene or operon
E. All of the above

18. (2 pts) The lac inducer enables transcription by binding to

A. Operator
B. Activator
C. Repressor
D. Lactose
E. RNA polymerases

19. (2 pts) The existence of polar mutations was one of the first lines of evidence that the three genes of the lac operon (lacZ, lacY, and lacA) were expressed as a single transcript. In a polar mutation

A. The lac operon is constitutive
B. The lac operon is uninducible
C. Nonsense alleles of lacZ reduce the expression of lacY
D. The lac permease protein has reversed polarity and exports lactose from the cell
E. Transcription of the lac operon is terminated before the lacZ gene is transcribed

20. (2 pts) In E. coli, the corepressor of tryptophan biosynthesis is

A. Tryptophan attenuator
B. Tryptophan hydroxylase
C. tRNA^Trp
D. Tryptophan
E. TRAP

21. (2 pts) The expression of the trp operon in E. coli is regulated by the trp repressor, but also by another regulatory mechanism called attenuation. Which of the following statements about attenuation is false?

A. The mechanism of attenuation involves the formation of stem-loop structures in the transcript.
B. The mechanism of attenuation requires translation of the transcript before transcription of the trp operon is complete.
C. Attenuation negatively regulates expression of the trp operon using a RNA-binding protein that controls alternative splicing.
D. Attenuation is a mechanism that causes the termination of transcription of the trp operon before the coding sequence of the first biosynthetic enzyme is reached.
E. Attenuation blocks the transcription of most of the trp operon in the presence of high concentrations of charged tRNA^Trp.

22. (2 pts) The Sex-lethal (Sxl) gene functions early in Drosophila development to control sex determination and dosage compensation. Which of the following statements is false?

A. SXL protein is present in females but absent in males
B. SXL protein negatively regulates the transcription of all five male-specific lethal genes involved in dosage compensation
C. The SXL protein is an RNA-binding protein
D. SXL protein inhibits translation of msl-2 mRNA
E. In the presence of SXL protein, several primary transcripts are spliced in a different way than in the absence of SXL protein

23. (2 pts) Women at high risk of breast and ovarian because of their genotype with respect to BRCA1 are

A. Homozygous for a gain-of-function allele of BRCA1
B. Heterozygous for a gain-of-function allele of BRCA1
C. Mosaic for a reciprocal translocation that places BRCA1 expression under the control of another gene
D. Homozygous for a loss-of-function allele of BRCA1
E. Heterozygous for a loss-of-function allele of BRCA1

24. (2 pts) Genes that control early development through their expression in the embryo are called

A. Male-specific lethals
B. Developmental early genes
C. Maternal-effect genes
D. Coordinate genes
E. Zygotic genes

25. (2 pts) Mutations in Drosophila that result in the transformation of one body segment into another are probably mutations in

A. Gastrulation genes
B. Homeotic genes
C. Segmentation genes
D. Execution genes
E. Maternal-effect genes

26. (2 pts) The enzyme that can be used to make a DNA copy of an RNA molecule is called

A. DNA polymerase I
B. Ligase
C. Primase
D. Reverse transcriptase
E. Topoisomerase II

Please see the lecture notes.

27. (2 pts) The best description of the Sanger method of sequencing DNA is

A. Labeled DNA molecules are fragmented by reactions that break the phosphodiester bond at specific bases, then sorted by size using electrophoresis
B. The products of four DNA synthesis reactions, each with a different density-labeled base, are analyzed by mass spectrometry
C. The products of four DNA synthesis reactions, each with a specific dideoxyribonucleotide mixed with the four dNTPs, are sorted by size using electrophoresis
D. DNA synthesis is carried out in a device that detects the proton released when a dNTP is incorporated into the growing chain
E. Single stranded DNA molecules are driven through a nanopore by a charge difference, with each base producing a characteristic change in the voltage measured

Please see the lecture notes.

28. (2 pts) The human Huntington Disease gene was located using restriction fragment length polymorphisms. A restriction fragment length polymorphism is

A. Variation in the DNA methylase that modifies chromosomal DNA at a restriction site so that the genomic DNA can no longer be cleaved at that site
B. Variation in the sequence of genomic DNA that results in the presence or absence of a restriction site at a particular place in the genome
C. Variation in a gene encoding a restriction enzyme so that the restriction enzyme has an altered recognition site
D. Variation of the sequence of a segment of genomic DNA that alters the conformation of a specific restriction fragment, affecting its electrophoretic mobility
E. An insertion or deletion of several hundred bases of DNA that alters the size of a restriction fragment

Please see the lecture notes.

29. (2 pts) Which of the following statements about the production of targeted mutations (knockouts) in mice is false?

A. Production of a targeted mutation in mice begins with the construction of a targeting construct in a bacterial plasmid vector.
B. A targeting construct contains part of a mouse gene with an exon interrupted with a selectable marker.
C. Targeting constructs are linearized and introduced into mouse embryonic stem cells by transformation.
D. Successful targeting of the gene in embryonic stem cells is first screened for using selectable markers, then confirmed by PCR.
E. A nucleus from an embryonic stem cell bearing the knockout mutation is transferred to an enucleated egg, which when implanted into a female develops into a mouse homozygous for the knockout mutation.

Please see the lecture notes.

30. (2 pts) Which of the following types of sequences makes up the largest fraction of the human genome?

A. Pseudogenes
B. Exons
C. Interspersed repeats derived from transposable elements
D. Microsatellites
E. Promoter sequences

Please see the lecture notes.

31. (2 pts) Linkage disequilibrium is

A. A violation of segregation that produces a deviation from the 1:1 ratio of gamete types expected from a heterozygote
B. A violation of independent assortment that causes genes on different chromosomes to appear to be linked
C. A sense of disorientation experienced when conducting linkage analysis
D. The occurrence of haplotypes in a population at frequencies other than that predicted by allele frequencies, such that one pair of haplotypes will be more frequent than expected, while the other pair of haplotypes will be less frequent than expected
E. The occurrence of homozygotes and heterozygotes in a population at frequencies other than that predicted by allele frequencies, such that heterozygotes will be more frequent than expected, while both classes of homozygotes will be less frequent than expected

Please see the lecture notes.

32. (2 pts) A genome-wide association study is

A. A measurement of the correlation of restriction fragment length polymorphisms in an individual genome using two different restriction enzymes
B. A measurement of the differences in phenotypes between identical and fraternal twins
C. A survey of a population to determine whether different medical conditions are correlated
D. A case-control study in which a group of individuals with an inherited condition is compared to a group of controls at thousands of SNPs spanning the entire genome
E. A study in which microarrays are used to measure the expression of a large number of genes

Please see the lecture notes.

33. (2 pts) In the ENCODE project, the entire human genome was surveyed for DNase I hypersensitive sites. In chromatin, DNase I hypersensitive sites are associated with

A. Interspersed repetitive elements derived from mobile elements
B. Satellite DNA
C. Enhancers
D. Introns
E. Pseudogenes

Please see the lecture notes, as well as these lecture notes.

34. (2 pts) In the ENCODE project, total mRNA from cell lines and tissues was sequenced using a technique called RNA-seq. RNA-seq allows the discovery of

A. Enhancers
B. Sites in the genome that are heterochromatic
C. Interspersed repetitive elements derived from mobile elements
D. Sites in the genome associated with post-translational histone modifications like methylation or acetylation
E. Splice sites through the detection of junction reads when the sequence of an mRNA is aligned to the genome assembly

Please see the lecture notes.

35. (2 pts) In the ENCODE project, the entire human genome was surveyed using a technique called ChIP-seq, for chromatin immunoprecipitation followed by DNA sequencing. ChIP-seq allows the discovery of

A. Sites in the genome associated with a specific transcription factor for which an antibody is available
B. Sites in the genome associated with post-translational histone modifications like methylation or acetylation
C. Sites in the genome associated with RNA polymerase
D. All of the above
E. None of the above

Please see the lecture notes.

36. (5 pts.) Red-green colorblindness is an X-linked trait in humans. A man with normal color vision is married to a woman with normal vision whose father was colorblind. They have a daughter with Turner Syndrome (45, X0) who is colorblind. The best explanation for this is

A. Nondisjunction at meiosis I in the mother
B. Nondisjunction at meiosis I in the father
C. Nondisjunction at meiosis II in the mother
D. Nondisjunction at meiosis II in the father
E. Nondisjunction at an unknown division in the father

37. (5 pts) Which of the following is true about eukaryotic mRNA?

A. Processing of the nascent mRNA may begin before its transcription is complete.
B. Many RNAs can be transcribed simultaneously from one DNA template.
C. Termination occurs via a hairpin loop or use of rho factor.
D. Sigma factor is essential for correct initiation of transcription.
E. Processing occurs in the cytoplasm.

38. (5 pts) Which of the following describes an experimental strategy used to decipher the genetic code?

A. Analyzing the sequence of RNAs produced from known DNA sequences
B. Comparing the amino acid sequence of proteins to the nucleotide sequence of their genes in humans
C. Examining the polypeptides produced when RNAs of known sequence were translated
D. Analyzing mutants that changed the code

39. (5 pts) The bicoid (bcd) gene of Drosophila is known to encode a morphogen, a protein whose concentration affects the developmental fate of cells in the embryo. One of the key experiments leading to this conclusion is

A. DNA sequencing of the bcd gene reveals that it is a transcription factor
B. RNA in situ hybridization shows that the bcd gene is transcribed in the zygote during development
C. RNA in situ hybridization shows that the bcd gene is transcribed in the mother during oogenesis
D. Varying the number of copies of the bcd gene in the mother affects the position of the cephalic furrow in developing embryos
E. Embryos homozygous for loss-of-function alleles of the bcd gene develop abnormally

40. A rare human disease affected a family as shown in the pedigree below.

a. (5 pts.) The most likely mode of inheritance is

A. Autosomal dominant
B. Autosomal recessive
C. X-linked dominant
D. X-linked recessive
E. There is not enough information to solve the problem.

Explanation: Notice that there are no affected individuals with two unaffected parents. This tells us that the variant is dominant.

Affected females (II-2 and II-5) have unaffected sons (III-1 and III-6). An affected male (II-7) has an unaffected daughter (III-8). An affected male (II-11) has an affected son (III-12). This proves that this dominant variant cannot be X-linked; it must be autosomal.

b. (5 pts.) Individual 13 in generation III is given as a diamond shape rather than as a square or circle because the sex is not stated. You are a genetic counselor advising the couple (individuals 11 and 12 in generation II) on the probability that their next child (individual 13 in generation III) will be affected. The best answer to give them is

A. Your next child has a 25% chance of being affected.
B. Your next child has a 50% chance of being affected.
C. If your next child is a son, there is a 100% chance that he will be affected; if your next child is a daughter, there is a 50% chance that she will be affected.
D. If your next child is a daughter, there is a 100% chance that she will be affected; if your next child is a son, there is a 50% chance that he will be affected.
E. It is not possible to know the probability that your next child will be affected.

41. A rare human disease affected a family as shown in the pedigree below.

a. (5 pts.) The most likely mode of inheritance is

A. Autosomal dominant
B. Autosomal recessive
C. X-linked dominant
D. X-linked recessive
E. There is not enough information to solve the problem.

Explanation: There are no affected individuals with two unaffected parents. This shows that the variant is dominant.

An affected man (I-1) has three unaffected sons (100%) and three affected daugthers (100%). Another affected man (III-3) has four unaffected sons (100%) and four affected daughters (100%). Affected females (II-2, II-7, and III-10) have 50% affected children regardless of sex.

We see a total of seven affected daughters whose fathers are affected, and no unaffected daughters from these fathers. If this is an X-linked dominant, we would expect those results 100% of the time. If this is an autosomal dominant, we would expect seven affected and zero unaffected daughters only (1/2)⁷ or in 1/128 of such samples.

There are seven sons of affected fathers, all unaffected. If this is an X-linked dominant, we would expect those results 100% of the time. If this is an autosomal dominant, we would expect seven unaffected and zero affected sons only (1/2)⁷ or in 1/128 of such samples.

Taken together, the evidence strongly supports the model of an X-linked dominant. We would expect these results from an autosomal dominant only (1/128)(1/128) or in 1/16,384 of such samples.

b. (5 pts.) Individual 9 in generation IV is given as a diamond shape rather than as a square or circle because the sex is not stated. You are a genetic counselor advising the couple (individuals 3 and 4 in generation III) on the probability that their next child (individual 9 in generation IV) will be affected. The best answer to give them is

A. Your next child has a 25% chance of being affected.
B. Your next child has a 50% chance of being affected.
C. If your next child is a son, there is a 100% chance that he will be affected; if your next child is a daughter, there is a 50% chance that she will be affected.
D. If your next child is a daughter, there is a 100% chance that she will be affected; if your next child is a son, there is a 0% chance that he will be affected.
E. It is not possible to know the probability that your next child will be affected.

42. Phenylketonuria is a rare human disease inherited as an autosomal recessive. Phenylketonuria affected a family as shown in the pedigree and diagram of some of the results of a Southern blot below. The probe used in the Southern blot hybridizes to a portion of the phenylalanine hydroxylase (PAH) gene that is responsible for phenylketonuria, and detects a restriction fragment length polymorphism with two alleles: 8 kb and 4 kb.

a. (5 pts) Individual II-3 is affected. The expected molecular phenotype of individual II-3 is

A. A single band at 8 kb
B. A single band at 4 kb
C. A band at 8 kb and a band at 4 kb
D. Either A or C
E. Either B or C

Explanation: Individuals I-1 and I-2 are both heterozygous for PKU. They are also both heterozygous for the RFLP. Individual II-1 is not affected by PKU and is homozygous for the 8 kb RFLP. Individual II-2 is heterozygous for the RFLP and for PKU. We expect individual II-3, who is homozygous for PKU, to be homozygous for the 4 kb allele of the RFLP.

b. (5 pts) Individual II-4 is unaffected. The expected molecular phenotype of individual II-4 is

A. A single band at 8 kb
B. A single band at 4 kb
C. A band at 8 kb and a band at 4 kb
D. Either A or C
E. Either B or C

Explanation: Individual II-4 is either homozygous normal for PKU, like individual II-1, or heterozygous for PKU, like individual II-2. Individual II-4 can therefore either be homozygous for the 8 kb allele of the RFLP or heterozygous for the 8 kb and the 4 kb allele of the RFLP.

c. (5 pts) Individual II-2 marries a woman who is known to be a carrier of phenylketonuria. You are a genetic counselor advising the couple on the chance that their first child will be affected. The best answer to give them is

A. Your first child has a 25% chance of being affected.
B. Your first child has a 50% chance of being affected.
C. If your first child is a son, there is a 100% chance that he will be affected; if your first child is a daughter, there is a 50% chance that she will be affected.
D. If your first child is a daughter, there is a 100% chance that she will be affected; if your first child is a son, there is a 0% chance that he will be affected.
E. It is not possible to know the probability that your first child will be affected.

43. For this question on the lac operon, single letters are used to indicate the elements of the lac operon as follows:

I = lacI (lac repressor), I^- = loss-of function allele, I^S = superrepressor
P = lacP (lac promoter), P^- = loss-of-function allele
O = lacO (lac operator), O^C = operator-constitutive allele
Z = lacZ (beta-galactosidase), Z^- = loss-of-function allele
Y = lacY (lac permease), Y^- = loss-of-function allele

IPTG = isopropylthiogalactopyranoside, gratuitous inducer

In the strains listed below, indicate whether lacZ and lacY expression is constitutive (+ in the presence and absence of inducer), uninducible (- in the presence and absence of inducer), or inducible (+ in the presence of inducer, - in the absence of inducer). The first line shows a wild-type bacterium filled in as an example.