Exam 3 (11/12/2013)

1. (2 pts) Mutations are

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

Answers seen: A (9), B (19), C (3), D (27), E (0)

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

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

Answers seen: A (4), B (12), C (39), D (2), E (2)

Please see the lecture notes.

3. (2 pts) In conjugation

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

Answers seen: A (3), B (3), C (13), D (11), E (29)

Please see the lecture notes.

4. (2 pts) Which of the following statements is false regarding Hfr transfer?

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

Answers seen: A (15), B (5), C (6), D (32), E (1)

Please see the lecture notes.

5. (2 pts) A positively regulated genetic system

A. Is derepressed by inducer
B. Is more common in prokaryotes
C. Must override an inhibitor
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

Answers seen: A (7), B (2), C (3), D (29), E (18)

Please see the lecture notes.

6. (2 pts) A negatively regulated genetic system

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

Answers seen: A (3), B (25), C (21), D (0), E (10)

Please see the lecture notes.

7. (2 pts) Which enzyme cleaves lactose to glucose and galactose?

A. Glucose-6-phosphate dehydrogenase
B. Glucose synthetase
C. Beta-galactosidase
D. Phenylalaine hydroxylase
E. Alpha-glucosidase

Answers seen: A (0), B (0), C (58), D (0), E (0)

Please see the lecture notes.

8. (2 pts) Constitutive gene expression refers to

A. Constant, unregulated expression
B. Polycistronic mRNA
C. Coupled transcription/translation
D. Mutant repressor that no longer binds to inducer
E. Gene expression following the addition of inducer

Answers seen: A (52), B (0), C (3), D (4), E (2)

Please see the lecture notes.

9. (2 pts) The product of the lacI gene

A. Is cis-dominant
B. Induces lac operon transcription
C. Binds to the lac operator
D. Transports lactose into the cell
E. Controls splicing of lac pre-mRNA

Answers seen: A (0), B (8), C (45), D (3), E (3)

Please see the lecture notes.

10. (2 pts) The lac inducer enables transcription by binding to

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

Answers seen: A (2), B (14), C (21), D (10), E (12)

Please see the lecture notes.

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

Answers seen: A (11), B (7), C (11), D (10), E (19)

Please see the lecture notes.

12. (2 pts) In E. coli, the corepressor of tryptophan biosynthesis is

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

Answers seen: A (17), B (27), C (1), D (14), E (0)

Please see the lecture notes.

13. (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. 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.
B. Attenuation blocks the transcription of most of the trp operon in the presence of high concentrations of charged tRNA^Trp.
C. The mechanism of attenuation involves the formation of stem-loop structures in the transcript.
D. Attenuation negatively regulates expression of the trp operon using a RNA-binding protein that controls alternative splicing.
E. The mechanism of attenuation requires translation of the transcript before transcription of the trp operon is complete.

Answers seen: A (5), B (2), C (1), D (22), E (29)

Please see the lecture notes.

14. (2 pts) Regions of eukaryotic chromosomes with a low density of genes and large number of repetitive sequences are called

A. Euchromatin
B. Chromatids
C. Heterochromatin
D. Middle-repetitive sequences
E. Chromocenter

Answers seen: A (6), B (1), C (49), D (3), E (0)

Please see the lecture notes.

15. (2 pts) Regions of eukaryotic chromosomes that contain most of the functional genes are called

A. Euchromatin
B. Chromatids
C. Heterochromatin
D. Middle-repetitive sequences
E. Chromocenter

Answers seen: A (55), B (1), C (2), D (0), E (1)

Please see the lecture notes.

16. (2 pts) Which histone protein is present as a monomer within the nucleosome?

A. H1
B. H2A
C. H2B
D. H3
E. H4

Answers seen: A (49), B (1), C (1), D (3), E (5)

Please see the lecture notes.

17. (2 pts) Polytene chromosomes

A. Divide at the same rate as other chromosomes
B. Are large DNA molecules entirely composed of heterochromatin
C. Are present in all Drosophila cells
D. Form by repeated rounds of DNA replication without cell division
E. Do not contain sex chromosomes

Please see the lecture notes.

Answers seen: A (1), B (10), C (7), D (40), E (1)

18. (2 pts) The aggregation of centromeres and heterochromatin in polytene chromosomes is called

A. Nucleosome
B. Chromocenter
C. Chromatid
D. Chromosome
E. Chiasma

Please see the lecture notes.

Answers seen: A (8), B (45), C (2), D (0), E (3)

19. (2 pts) Specific enzymes modify histone tails post-translationally in order to affect the state of chromatin. Which of the following modifications is not made to histone tails?

A. Methylation
B. Acetylation
C. Farnesylation
D. Phosphorylation

Answers seen: A (1), B (3), C (51), D (3)

Please see the lecture notes.

20. (2 pts) There is an inversion of the X chromosome in Drosophila with one breakpoint in euchromatin near the white (w) gene and the other breakpoint in the pericentric heterochromatin of the X chromosome. Flies carrying this inversion show variegated expression of the normal allele of the white gene, that is, their eyes are a mottled mixture of red and white. Position-effect variegation is caused by

A. Instability of DNA near the white gene, causing it to be deleted in some cells
B. Reversion of the inversion to the normal X chromosome sequence in some cells
C. Activation of apoptosis (programmed cell death) in the eye
D. Spreading of the heterochromatic state to normally euchromatic sequences near the rearrangement breakpoint in some cells
E. None of the above

Answers seen: A (4), B (5), C (0), D (45), E (5)

Please see the lecture notes.

21. (2 pts) Drosophila males (XY) are killed by the recessive male-specific lethal mle, but females are not affected. Drosophila chromosomal females (XX) are transformed into sterile males by the recessive sex-transforming mutation tra, but males are not affected. Drosophila chromosomal females carrying both of these mutations (XX; mle/mle; tra/tra) are viable males because

A. In the absence of TRA protein, the X chromosome is inactivated
B. In the absence of MLE protein, the X chromosome is inactivated
C. In the absence of TRA protein, chromosomal females are transformed into males, but dosage compensation is not activated
D. In the absence of TRA protein, the mle primary transcript is spliced to form an mRNA that encodes a nonfunctional product
E. In the absence of MLE protein, the X chromosome is hypertranscribed

Explanation: A. The X chromosome is not inactivated in Drosophila.
B. The X chromosome is not inactivated in Drosophila.
D. If mle were inactivated in the absence of TRA (that is, in males), dosage compensation would not occur. We know that TRA is not active in males because tra/tra males are normal.
E. We know that MLE is required in males, because mle/mle males are inviable. The X chromosome is hypertrancribed in the presence of MLE.

Answers seen: A (1), B (4), C (41), D (6), E (7)

Please see the lecture notes.

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

SXL protein, an RNA-binding protein, is not a likely candidate for a regulator of transcription. In addition, msl-2 is the only male-specific lethal that is regulated in females. Blocking the splicing and translation of the msl-2 transcript keeps the entire dosage compensation complex from assembling, and keeps dosage compensation off in females.

Answers seen: A (1), B (16), C (1), D (31), E (9)

Please see the lecture notes.

23. (2 pts) Retinoblastoma is caused by a defect in the RB1 gene. The RB1 gene is an example of

A. A tumor-suppressing gene
B. A gene controlling mental development
C. A tumor-causing reciprocal translocation
D. An oncogene
E. A viral oncogene

Answers seen: A (54), B (0), C (2), D (3), E (0)

Please see the lecture notes.

24. (2 pts) RAS is a small GTPase that is active when bound to GTP and inactive when bound to GDP. In many tumors, single amino acid substitutions activate RAS by eliminating its ability to hydrolyze GTP. The Ras gene is an example of

A. A tumor-suppressing gene
B. A gene controlling apoptosis
C. A receptor tyrosine kinase
D. An oncogene
E. A G-protein coupled receptor

Answers seen: A (8), B (1), C (3), D (47), E (0)

Please see the lecture notes.

25. (2 pts) Women at high risk of breast and ovarian because of their genotype with respect to BRCA1 are

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

Answers seen: A (15), B (36), C (2), D (5), E (1)

Please see the lecture notes.

26. (2 pts) Li-Fraumeni Syndrome is a hereditary cancer syndrome that places people at elevated risk of developing a number of different tumor types. Li-Fraumeni Syndrome is caused by a defect in the TP53 gene, which encodes the p53 protein. Mutations in TP53 increase the risk of cancer because

A. p53 is an enzyme in the excision repair pathway
B. p53 normally inhibits apoptosis
C. p53 normally increases the rate of cell division
D. p53 negatively regulates cell proliferation and positively regulates apoptosis in the presence of DNA damage
E. p53 is required for signal transduction from receptor tyrosine kinases

Answers seen: A (5), B (2), C (5), D (44), E (2)

Please see the lecture notes.

27. (2 pts) Genes expressed in the mother that are required for the development of the embryo are called

A. Zygotic genes
B. Developmental early genes
C. Gap genes
D. Pair-rule genes
E. Maternal-effect genes

Answers seen: A (0), B (0), C (0), D (0), E (59)

Please see the lecture notes.

28. (2 pts) Genes that control early development through their expression in the embryo are called

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

Answers seen: A (1), B (22), C (25), D (10), E (0)

Please see the lecture notes.

29. (2 pts) Mutations in Drosophila that result in the transformation of one body segment into another are probably mutations in

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

Answers seen: A (1), B (14), C (4), D (38), E (1)

Please see the lecture notes.

30. (2 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. RNA in situ hybridization shows that the bcd gene is transcribed in the mother during oogenesis
B. RNA in situ hybridization shows that the bcd gene is transcribed in the zygote during development
C. Varying the number of copies of the bcd gene in the mother affects the position of the cephalic furrow in developing embryos
D. DNA sequencing of the bcd gene reveals that it is a transcription factor
E. Embryos homozygous for loss-of-function alleles of the bcd gene develop abnormally

Explanation: A. Many genes are transcribed during oogenesis; they are not all morphogens.
B. Many genes are transcribed during zygotic development; they are not all morphogens.
D. There are many transcription factors; they are not all morphogens.
E. Mutant alleles of many genes affect embryonic development; they are not all morphogens.

Answers seen: A (4), B (8), C (33), D (0), E (13)

Please see the lecture notes.

31. 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.

Answers seen (correct, 4 pts/incorrect, 0 pts): A (48/11), B (53/8), C (41/18), D (49/10), E (55/5), F (47/12), G (46/13), H (44/15), I (28/31), J (41/18)

Please see the lecture notes.