1
Homework # 4
Please turn in answers on a scantron in class Monday, Feb. 18. Free scantrons handed out in class
and they are sold at the Bookstore
1. DNA replication is semiconservative because
a. DNA polymerase is socially liberal but fiscally responsible.
b. both parental strands serve as templates for replication.
c. only one parent strand serves as a template for replication.
d. promoters have consensus sequences.
2. In a deoxyribonucleotide, 5’ and 3’ refer to the
a. short pieces of RNA that initiate DNA replication.
b. carbons where (respectively) the pyrimidine and purine bases are attached.
c. carbons where (respectively) the phosphate and hydroxyl groups are attached.
d. the number of nucleosomes in homologous chromosomes.
3. One RNA primer is sufficient for initiating replication of the leading strand whereas a new primer
is needed for each Okazaki fragment on the lagging strand.
a. T
b. F
4. An RNA primer is required to initiate DNA replication because
a. DNA polymerase can only add nucleotides to a free 3' OH.
b. the RNA primer opens up the helix and makes single stranded DNA.
c. the RNA primer joins Okazaki fragments and remains in the daughter strands.
5. DNA is such an ideal genetic material because it is usually replicated with high fidelity, carries
information and it is capable of change.
a. T
b. F
6. A heritable change in DNA sequence is best described as:
a. Epigenetic event
b. Methylation event
c. Mutation
d. Chiasma
7. Mutations caused by DNA replication errors are relatively rare because
a. Some polymerases have exonuclease activity and can remove incorrect bases from the 3’
end of a developing strand
b. Helicases have endonuclease activity and remove incorrect bases at replication forks
c. Arsenic is only substituted for phosphorous under lacustrine conditions
8. In eukaryotes, DNA replication is initiated at the centromere and telomere of each chromosome
a. T
b. F
2
9. At a bidirectional replication fork, lagging strands are
a. Where Okazaki fragments occur
b. Where new strands are synthesized 3’ to 5’
c. Always on top
d. Responsible for selective abortion of floral organs
10. The loss of DNA at telomeres during mitosis is due to
a. Unequal sister strand crossovers
b. Telomerase gnawing at the ends of chromatids
c. DNA replication issues
d. Colchicine interfering with the formation of spindle fibers
11. A comparative analysis of the DNA sequence of the BAD2 genes of rice and maize reveals that
the two genes are so similar that they are hypothesized to have evolved from a common
ancestral gene. This is best described as an example of
a. ornithology.
b. syntney.
c. orthology.
d. Homology.
12. What process gives rise to new alleles?
a. recombination
b. translation
c. levitation
d. mutation
13. What process gives rise to new combinations of alleles at different loci?
a. recombination
b. translation
c. levitation
d. mutation
14. The RNA primers that initiate DNA replication are the same as those that initiate transcription.
a. T
b. F
15. Only one of the two DNA strands in the double helix serves as template for transcription
throughout the entire length of the chromosome and throughout the life of the organism.
a. T
b. F
16. Transcription of genes located on the 20 chromosomes of maize occurs
a. in the cytoplasm.
b. in the nucleus.
c. in both the nucleus and cytoplasm.
d. only in heterochromatin.
3
17. RNA is different from DNA in that
a. RNA is always single stranded whereas DNA is always double stranded.
b. RNA is transcribed 3’ → 5’.
c. RNA has the base uracil whereas DNA has the base thymine.
d. RNA is found only in prokaryotes
18. In eukaryotes, informational and functional RNAs are
a. synthesized by the same RNA polymerase.
b. formed in the cytoplasm.
c. coded for by genes in the nucleus.
d. maternally inherited.
19. A promoter is
a. translated as methionine.
b. the site of assembly of ribosomes.
c. a binding site for RNA polymerase.
d. a binding site for DNA polymerase.
20. What segment of a gene would be most likely to have the sequence TATA ~ 30 bp upstream (5’)
from the transcription start site?
a. Promoter
b. Coding region
c. Intron junction
d. Termination signal
21. The sequence of the promoter can be found in a mRNA sequence.
a. T
b. F
22. The synthesis of the mRNA stops when
a. a stop codon is encountered.
b. the telomere is reached.
c. the polymerase encounters the promoter.
d. a specific sequence in the DNA template is reached that serves as a signal for
termination.
23. mRNA processing in eukaryotes refers to
a. 5’ caps, 3’ tails, and intron removal.
b. 3’ caps, 5’ tails, and exon removal.
c. transport of the mRNA to the cytoplasm.
d. coupling of the ribosomal subunits.
24. A diagnostic sequence always found in 5’ UTRs is the ATG start codon.
a. T
b. F
25. Exons are always longer than introns.
a. T
b. F
4
26. One of the explanations for the fact that in eukaryotes ~ 25,000 genes can specify ~ 100,000
proteins is
a. alternative removal of introns and splicing of exons.
b. alterative 5’ capping.
c. alternative use of 3’ and 5’ promoters.
d. each gene is transcribed 5’ to 3’ and 3’ to 5’ on leading and lagging strands.
27. Translation of the mRNAs transcribed from the genes located on the 20 chromosomes of maize
occurs
a. in the cytoplasm.
b. in the nucleus.
c. in both the nucleus and cytoplasm.
d. only in homozygotes
28. Transfer RNAs (tRNAs)
a. are very specialized, with each tRNA able to carry a specified amino acid.
b. very generic, with each tRNA able to carry any of a number of amino acids.
c. are coded for by sequences in introns.
d. consists of 15S and 35S subunits.
29. The point of attachment of an amino acid to a tRNA is called the anticodon.
a. T
b. F
30. Ribosomes are
a. specific to each type of mRNA.
b. coded for by nuclear genes.
c. formed by exon shuffling.
d. double stranded RNA molecules.
31. The rRNA subunits in ribosomes are transcribed from DNA, but are not translated.
a. T
b. F
32. In the process of translation, the ribosome moves
a. 5’ to 3’ on the mRNA.
b. 3’ to 5’ on the mRNA.
c. counterclockwise around the tRNA.
d. 5’ to 3’ on the antisense DNA.
33. During the elongation phase of the translation process, the next incoming tRNA will
a. arrive at the ribosome’s P site.
b. carry an amino acid at its 5’ end.
c. have an anti-codon sequence complementary with the codon in the mRNA.
d. arrive at the ribosome’s A site.
34. The mechanism by which a stop codon stops translation is that
a. tRNAs with affinity to the stop codon always bring methionine.
b. polymerases have exonuclease activity.
c. there are no tRNAs with anticodons matching the stop codon.
d. the ribosomes pinch off the completed sequence of amino acids.
5
35. The DNA code is degenerate because
a. the same codons specify different amino acids in different organisms.
b. start and stop codons are reversed in prokaryotes and eukaryotes.
c. the same codon can specify different amino acids.
d. the same amino acid can be specified by more than one codon.
Given the following DNA sense strand – 5’ ATG GCC TGG ACT TCA 3’
36. Which is the corresponding DNA anstisense strand?
a. 3’ TAC CGG ACC TGA AGT 5’
b. 5’ TAC CGG ACC TGA AGT 3’
c. 3’ UAC CGG ACC UGA AGU 5’
d. 5’ UAC CGG ACC UGA AGU3’
37. Which is the corresponding mRNA?
a. 3’ UAC CGG ACC UGA AGU 5’
b. 5’ UAC CGG ACC UGA AGU 3’
c. 5’ AUG GCC UGG ACU UCA 3’
d. 3’ AUG GCC UGG ACU UCA 5’
38. Which is the correct translation?
a. Ser Thr Trp Ala Met
b. Tyr Arg Thr Stop Ser
c. Met Ala Trp Thr Ser
d. Thr Ser Gly Pro Val
39. If you know the amino acid sequence of a polypeptide you can deduce the exact DNA code of the
corresponding gene.
a. T
b. F
40. In “The gene for fragrance in rice” paper, the authors conclude that the functional basis for the
difference between aroma vs. non-aroma alleles is probably due to a premature stop codon in the
non-aromatic allele due to a frameshift mutation. This causes aroma to be dominant and lack of
aroma to be recessive.
a. T
b. F