University of California, Davis

BIS2A Section B- Britt, Spring 2012 Midterm 3

1)Which of the following statements regarding meiosis I is false?

1. 1. Crossing-over occurs between chromosome arms.
   2. Meiosis I involves the separation of chromosomes into daughter cells.
   3. Synapsis occurs during meiosis I.
   4. At the end of meiosis I, all chromosomes have one chromati
   5. DNA double strand breaks are induced in all chromosomes.

2. Sexual reproduction increases genetic diversity in three ways. Which of the mechanisms described below is not one of these ways?
   1. New combinations of alleles on a single chromosome are created by crossing over.
   2. New combinations of alleles are created via increased mutation rate
   3. New combinations of alleles are created via fertilization
   4. New combinations of alleles are created via segregation of homologs to the poles

3. Elephants and mosses differ in many interesting ways. Strangely, the only difference specifically described in this course is the fact that:
   1. After fertilization but before meiosis, elephant cells undergo many mitoses but moss cells don’t.
   2. After fertilization but before meiosis, moss cells undergo many mitoses but elephant cells don’t.
   3. After meiosis but before fertilization, elephant cells undergo many mitoses but moss cells don’t.
   4. After meiosis but before fertilization, moss cells undergo many mitoses but elephant cells don’t.

4. Avery, MacLoed, and McCarty were able to genetically transform “rough” pneumococcus to “smooth”, using an extract from killed pneumococci, They concluded that DNA rather than protein was the genetic material because:
   1. The extract, when treated with protease, still transformed cells
   2. The extract, when treated with protease, no longer transformed cells
   3. The extract, when treated with DNAse, still transformed cells
   4. The extract, when treated with DNase, no longer transformed cells
   5. Both a and d

5. Telomerase solves the “linear chromosome end problem” by…
   1. Extending the 3’ end of the chromosome
   2. Extending the 5’ end of the chromosome
   3. Providing a primer for the initiation of replication
   4. Sealing the gap between Okazaki fragments
   5. Both c and d

6. The majority of the DNA in a bacterial chromosome has been synthesized by
   1. Telomerase
   2. DNA polymerase III
   3. DNA polymerase I
   4. DNA ligase
   5. Primase

7. Adding another nucleotide to the 3’ of a growing DNA chain is energetically favorable because…
   1. You are making a larger molecule out of smaller components
   2. You are performing a condensation reaction, and releasing a water molecule
   3. You are releasing two inorganic phosphates
   4. You are releasing a single inorganic phosphate
   5. It’s NOT an energetically favorable reaction.

8. The plant Arabidopsis has duplex DNA. This DNA is extracted, hydrolyzed to completion, and the number of As, Cs, Gs, and Ts are accurately quantified. Which of the following might be a possible result?

a. 10% A, 20% C, 40% G and 30% T.

b. 10% A, 10% C, 40% G and 40% T.

c. 20% A, 20% C, 30% G and 30% T.

d. 20% A, 30% C, 30% G, and 20% T.

9. One DNA strand has the sequence 5’ACGTGTGC 3’. The complementary strand will have the sequence:
   1. 5’ACGTGTGC 3’
   2. 5’TGCACACG 3’
   3. 5’GCACACGT 3’
   4. 5’GCACACGU 3’
   5. 5’ACGUGUGC 3’

10. Newly synthesized DNA is illustrated here. The Okazaki fragments have not yet been processed. Which of the Okazaki fragments was synthesized first?

1. 

   The one on the right
2. The one on the left
3. The one in the middle
4. There’s no way to tell.

11. Mismatched bases that escape the notice of DNA polymerase can still be fixed via “mismatch repair”. This process works by…

1. Recombination
2. Recognizing the mismatched bases and ripping out the older strand
3. Recognizing the mismatched bases and ripping out the younger strand
4. Recognizing the damaged strand, and ripping it out
5. Mismatches can’t be repaired- there’s no way to tell which base is incorrect

12. Mutations can be repaired by…

1. Recombination
2. Recognizing the mismatched bases and ripping out the older strand
3. Recognizing the mismatched bases and ripping out the younger strand
4. Recognizing the damaged strand, and ripping it out
5. Mutations can’t be repaired- there’s no way to tell which base pair is incorrect

13. Chromosomes are made of both DNA and protein, and were known to contain the heritable material. However, no one knew if the DNA or the protein was actually the chemical that encoded “inheritance”. It was very hard to convince people that DNA, rather than protein made up the genetic code because:
    1. Proteins have a branching structure, while DNA has a linear structure
    2. Proteins have 20 different “doo-dads” (R groups) while DNA only has 4 (bases)
    3. DNA has no enzymatic activities, while proteins do.
    4. a and b
    5. b and c

14. Transcription is like DNA replication in that:

1. Synthesis is antiparallel to the template strand
2. A promoter is required for initiation of synthesis
3. Synthesis of the leading and lagging strand are coupled, physically
4. dNTPs are required
5. c and d

15. Which of the following activities does tRNA aminoacyl synthase perform?

1. Error-checking for correctly charged tRNAs
2. Formation of a peptide bond
3. Charging of tRNAs with amino acids
4. Progression from 5’ to 3’ along the transcript
5. a and c

16. A nonsense mutation is …

1. When the two bases in the double helix don’t match
2. When the sequence of a codon is changed, but its meaning is not.
3. A mutation that changes the amino acid encoded
4. A frameshift mutation
5. A mutation to a stop codon

17. In the diagram below, which DNA strand will act as the template strand for RNA polymerase?

1. The top one
2. The bottom one
3. both strands will be copied

4. Its impossible to tell from this diagram

18. In eukaryotes, the correct reading frame for translation is precisely determined by:

1. The position of the 5’ cap
2. The position of the most 5’ methionine codon
3. The position of the most 3’ methionine codon
4. The position of a methionine codon in the center of the gene
5. The position of the stop codon

19. Eyes are useless in a lightless environment, and many cave-dwelling animals have lost their eyes. This loss might be due to…

1. Their failure to use their eyes
2. Genetic drift
3. Natural selection for eyelessness
4. b and c are possible

20. The peacock’s tail is an example of…

1. Genetic drift
2. Natural selection
3. Sexual selection
4. Artificial selection
5. Inbreeding

21. A very short eukaryotic mRNA has the sequence…

5’ ACGCCGAUCAUGGGCAUGCGAGUAUAAAAACUGG 3’

How many amino acids are encoded by this message?

1. 11
2. none
3. 4
4. 7
5. 5

22. Some Labrador Retriever dogs are solid black, and some are brown. A black male is crossed with a brown female, resulting in 12 “F1” puppies (6 male and 6 female)- all of which are black. Since this is an unscrupulous “puppy mill” breeder, she proceeds to sibling-cross the puppies, producing 200 F2 pups. 145 of these pups are black, and 55 are brown. The most likely explanation for these observations is:

1. The brown phenotype (appearance) is due to a recessive mutation.
2. The brown allele is transmitted more rarely than the black allele.
3. There are more black genes than brown ones.
4. The “brown” allele is dominant to the “black” allele.
5. The gene for “Brown” is on the X chromosome.

23. The mutant allele coding for orange coat color in cats is located on the X chromosome. Cats heterozygous for this allele (X Xo cats) have patches of orange mixed with patches of black. They’re called “torties”, which is short for “tortoise shell”. An orange male is crossed to a tortie female. What fraction of their kittens are predicted to be orange?