Fall 2004 Animal Science 213 Animal Genetics EXAM 1 125 Points Exam MUST BE completed IN INK!
NAME:___Key______
Multiple Choice (4 points each)
1. Meselson and Stahl grew E. coli in medium containing radioactive, “heavy” 15N for many generations, then switched the medium to non-radioactive, “light” 14N. After the first generation that grew in the presence of 14N, they separated the resulting DNA by centrifugation and observed one band of intermediate weight (14N/15N). What could they conclude? a. XDNA replication was not conservative b. DNA replication was dispersive c. DNA replication was conservative d. none of the above
2. After the second generation, Meselson and Stahl observed two equal bands of DNA, one entirely “light” 14N, the other intermediate, 14N/15N. After the third generation, the band of 14N was bigger, the band of intermediate 14N/15N was smaller. Through subsequent generations, the intermediate band continued to decrease in proportion, but never entirely disappeared. What could the researchers conclude? a. DNA replication was dispersive b. XDNA replication was semi-conservative c. DNA replication was not semi-conservative d. none of the above
3. During mitotic anaphase, each sister chromatid a. lines up on the metaphase plate b. attaches itself to spindle fibers c. Xbegins to move toward opposite poles d. none of the above
4. Gametes are ______. a. Xhaploid b. diploid c. both a and b d. neither a nor b
1 5. What is the TOTAL number of hydrogen bonds that would form between the DNA strand below and its complement? (If it helps you, write in the complement strand)
5’AAGCAT3’
a. 16 b. 12 c. X14 d. 6
6. In E coli, which have the lac operon, when lactose is present, where is Lac I? a. bound to the promoter of the lac operon b. bound to the operator of the lac operon c. Xfree in the cytoplasm d. none of the above
7. When Lac I is bound to the lac operon, can RNA polymerase form the initiation complex? a. Yes b. XNo c. Sometimes d. Yes, but only after a $20 bribe
8. The genetic code is degenerate. This means a. there is “spillover” such that different codons can designate more than one amino acid b. Xthere is more than one codon designating a single amino acid c. different species use different genetic codes d. the code is not universal
9. When a ribosome is moving along an mRNA molecule and encounters a stop codon, which of the following occurs? a. No tRNA binds the A site b. A tRNA bearing the termination amino acid binds the A site c. The ribosome subunits dissociate d. Xa and c e. b and c
10. If an animal has a diploid number of 20, how many chromosomes will be visible at the end of mitotic prophase? a. 10 b. 5 c. X20 d. chromosomes are not visible at this point
2 11. When a nucleotide is added to a growing DNA or RNA polymer, what is removed? a. Nothing; the nucleotide is added intact b. One base c. One phosphate d. XTwo phosphates
12. During synthesis of a growing DNA or RNA polymer, the new nucleotide is a. Xattached by the 5’ phosphate of the new nucleotide to the 3’ hydroxyl of the last nucleotide in the polymer b. attached by the 5’ carbon of the new nucleotide to the 3’ hydroxyl of the last nucleotide of the polymer c. attached by the 5’phosphate of the last nucleotide in the polymer to the 3’ hydroxyl of the new nucleotide d. attached by the 5’ phosphate of the first nucleotide in the polymer to the 3’ hydroxyl of the new nucleotide
13. Which of the following is the universal start codon? a. XAUG b. GUA c. UAC d. AAA
14. For the following codon 5’UAC3’, what would be the anticodon on the tRNA? a. 5’AUG3’ b. X3’AUG5’ c. 5’UAC3’ d. none of the above
15. During meiosis, there is no assignment of paternal and maternal chromosomes to one daughter cell or the other. This is called a. random variation b. Xindependent assortment c. recombination d. none of the above
16. Below is a portion of an mRNA strand. What would be the sequence of the template DNA strand?
mRNA: 5’AUGUCCCUAGCC3’
a. 5’UACAGGGAUCGG3’ b. 5’TACAGGGATCGG3’ c. 3’UACAGGGAUCGG5’ d. X3’TACAGGGTACGG5’
3 17. A nonsense mutation is an alteration that a. produces a codon that will recruit a tRNA for a non-essential amino acid. b. Xresults in a stop codon. c. causes the ribosome to insert an extra amino acid. d. none of the above.
Use the following mRNA sequence and codon usage table for the next three questions (4 points each).
mRNA sequence: 5’AUGUGUACUGGUG
18. Write out the predicted amino acid sequence from the mRNA.
MET CYS THR GLY
19. If there were an A inserted after the second G (underlined), what would be the predicted amino acid sequence?
MET STOP or no protein/polypeptide produced
20. The following has TWO parts. The mRNA sequence above is from the 5’ end of the CAP gene, a regulator of the lac operon. In the face of the insertion mutation described above, would the gene be transcribed at relatively high or low levels if the bacteria were grown 1) in the presence of lactose? 2) In the presence of glucose and lactose? 1) Low; 2) Low
4 21. Okazaki fragments are formed because a. DNA is cleaved by exonucleases prior to replication b. DNA is synthesized continuously c. DNA is replicated conservatively d. XDNA is replicated semi-discontinuously
22. True/False (1 point each)
__T___ Mitosis preserves the genetic composition of daughter cells.
__F___ DNA replicates between Meiosis I and Meiosis II
__T___ During mitosis, one sister chromatid from each chromosome pair is pulled to each pole.
__F__ Hydrogen bonds form between the sugar residues of complementary DNA strands.
__T___ The 5’ cap and 3’ poly-A tail increase stability of eukaryotic mRNA.
__F___ The four haploid spermatids that result from meiotic division of a primary spermatocyte are all genetically identical.
__T__ The difference between polar bodies and the oocyte is in the amount of cytoplasm.
__T___ Primary spermatocytes and oocytes are diploid.
___T__ The two strands of a linear, double-stranded DNA molecule are complementary, not identical.
Sentence Completion (2 points per blank or completion)
23. A tetrad consists of __4__ (number) chromatids. After Meiosis I, the tetrad has divided into a dyad consisting of __2__(number) chromatids. After Meiosis II, the dyad has divided into a monad consisting of __1__ (number) chromatids.
24. Two sources of variation arising from meiosis are _recombination______and __independent assortment______.
25. In a non-dividing cell, DNA is uncoiled and dispersed. This structural form of DNA is referred to as _chromatin______.
26. In a dividing cell, DNA is condensed and coiled into structures called _chromosomes______.
27. In eukaryotes the structure responsible for protein synthesis is the _ribosome______.
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28. DNA methylation increases / decreases (circle the right answer) transcription efficiency.
29. The enzyme responsible for DNA synthesis is DNA polymerase______.
30. (6 points) The following has TWO parts. 1) Underneath each structure, designate which of the two structures below is a purine and which is a pyrimidine and 2) assign each of the four nucleotide bases to its appropriate base structure.
Thymine or Uracil Adenine
Cytosine Guanine
______Pyrimidine ______Purine
31. (4 points) The following has TWO parts. Below is a newly opened replication fork. 1) draw an arrow showing the direction in which each new strand of DNA would be synthesized during replication. 2) label the strand orientation (5’ and 3’) of each new strand. 3’ 5’
5’ 3’ 3’ 5’
3’ 5’
32. (2 points) Below is a newly transcribed RNA molecule composed of introns and exons
Draw the predicted mRNA if splicing occurred normally.
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Bonus (6 points)
Consider a bacterial operon that encodes the enzymes for synthesis of a theoretical amino acid called xenophan. The operon is regulated by a separate gene, R, deletion of which causes loss of xenophan synthesis. Under normal circumstances, when xenophan is present, no enzymes are made. Mutations in the operator region result in repression whether or not xenophan is present in the growth medium.
R Operator Promoter xen1 xen2
A.A.(2 points) Is the operon under negative or positive control? The operon is under positive control (deletion of the R gene causes loss of xenophan synthesis; therefore, R upregulates or turns on gene expression).
B. (2 points) Propose a function for R. R is an activator that turns on gene expression by facilitation formation of the RNA polymerase initiation complex.
C. (2 points) How might the presence of xenophan influence the expression of R? The presence of xenophan indicates that it is not necessary to synthesize more xenophan, so xenophan may decrease expression of R. Or it may bind to R to change its conformation such that it cannot bind the operator and turn up expression of the xenophan synthesis genes. Or it may induce expression of another regulator molecule that acts as a repressor of the xenophan operon (competitor of R).
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