Download Biology 303 EXAM II 3/16/99 NAME

Biology 303 EXAM II 3/22/01 NAME_______________________________
-----------------------------------------------------------------------------------------------------------------This exam consists of 40 questions worth 2.5 points each. On the separate answer form, please fillin the single best choice for each question. Be sure to fill-in your student number on the answer form. Good
luck!
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1. Which statement is true?
1. two genes on the same chromosome can never assort independently from one another.
2. two genes on different chromosomes will likely display linkage.
3. two genes on the same chromosome will always appear to be genetically linked to one another in a
dihybrid cross.
4. if two genes are genetically linked, the frequency of recombination between them will be less than 50%.
2. Assuming that genes "A" and "B" display complete linkage (that is, no recombination between them),
how many different phenotypes should appear among the offspring of the cross AaBb X aabb ?
1. 2
2. 1
3. 4
4. can't tell
3. Genes on the same chromosome that are more than 50 map units apart
1. produce only crossover gametes.
2. must reside on opposite sides of the centromere.
3. are too far apart to produce crossover gametes.
4. behave no differently than as if they were on different chromosomes.
4. The amount of recombination between two loci
1. increases when each locus is heterozygous.
2. can be estimated by self-fertilizing an organism and observing the offspring.
3. is a function of the physical distance between the loci.
4. is greater in somatic cells than germ cells.
5. Mapping experiments become less accurate when the distances between genes become large because
1. crossover gametes become less common.
2. multiple crossovers are more common.
3. interference is greater when the distance between genes is large.
4. recombination occurs less frequently in long chromosomes.
6. The frequency of crossover between genes A and B is 7.5%, and between B and C
is 34.5%, and between A and C is 41.4%. What is the order of these three genes?
1. B C A
2. A B C
3. C A B
4. It cannot be determined from these data.
7. If a mouse with the genotype AaBb is test-crossed, what proportion of offspring genotypes would
Mendel have expected?
1. 25% of each possible combination of traits: A and B, A and b, a and B, and a and b.
2. 9/16 would show A and B, 3/16 would show A and b, 3/16 would show a and B, and 1/16
would show a and b.
3. Half of the offspring showing both the A and B phenotypes and half showing both the a and b
phenotypes.
4. 3/4 would show A and B, while 1/4 would show a and b.
8. In the example above, what offspring would be expected if the two genes are 10 map units apart and the
heterozygote has the dominant alleles on one chromosome and the recessive alleles on the other?
1. 45% of the offspring will exhibit A and B, 45% will exhibit a and b, 5% will exhibit A and b,
and 5% will exhibit a and B.
2. 40% of the offspring will exhibit A and B, 40% will exhibit a and b, 10% will exhibit A and b,
and 10% will exhibit a and B.
3. 45% of the offspring will exhibit A and b, 45% will exhibit a and B, 5% will exhibit A and B,
and 5% will exhibit a and b.
4. 25% of each possible combination of traits: A and B, A and b, a and B, and a and b
9. A female fruit fly heterozygous for three linked mutant alleles a,b,c, (genotype AaBbCc) is crossed with
a male fly that is homozygous recessive for all three mutant alleles. If the phenotypes of the most common
offspring are ABc and abC, and the least common offspring are aBc and AbC, then the order of the genes a
b c on the chromosome is:
1. a b c
2. b a c
3. b c a
4. not enough information to tell.
10. It is usually easier to identify spontaneous mutations in bacteria than most eukaryotes because
1. mutations are visible as color changes within a bacterial plaque.
2. mutations can be induced via exposure to the bacteriophage T1.
3. given the haploid condition of most bacteria, mutations are typically expressed directly in the
descendant cells.
4. mutations occur at a much higher frequency in bacteria.
11. In conjugation, genetic information is transferred
1. from one cell into the culture medium, where it is taken up by another cell.
2. with the help of a viral go-between.
3. in a bidirectional fashion between two cells.
4. from one bacterium to another.
12. Hfr bacterial cells exhibit the highest levels of gene transfer because
1. they contain two distinct F factors.
2. they induce the production of significantly more sex pili than normal F+ cells.
3. rather than existing as an isolated plasmid, the F factor is incorporated within the
main chromosome where it can transfer many genes at one time.
4. they do not require conjugation for gene transfer.
13. In cotransformation
1. non-linked genes are passed simultaneously from one cell to another.
2. a bacterial cell receives two adjacent genes on a single piece of DNA from the medium.
3. two bacterial cells within a culture are transformed by the same genetic material.
4. it is not uncommon for the entire bacterial chromosome and F factor to be passed from one cell
to another.
14.
The genetic material of most living things has each of the following properties EXCEPT:
1. it encodes much information.
2. it never changes.
3. it has the ability to be replicated.
4. it has the capability of being mutated.
15. In Hershey and Chase's classic experiments with radioactively labeled phages
1. the sulfur contained in DNA was radiolabeled.
2. protein was demonstrated to be the genetic material.
3. the phosphorous of proteins was radiolabeled.
4. infection rather than transfection was used.
16. Avery, McLeod and McCarty had which of the following results?
1. DNase destroyed the transforming activity.
2. The transforming principle was too complex and difficult to be purified.
3. RNase destroyed the transforming activity.
4. Protease destroyed the transforming activity
17.
HIV, the AIDS virus, like other retroviruses
1. has a genome made of DNA.
2. uses reverse transcriptase to convert RNA to DNA.
3. neither a nor b.
4. both a and b.
18.
A difference between DNA and RNA is that
1. RNA is genetic material.
2. only DNA is involved in the expression of genetic information.
3. DNA has one extra oxygen on each ribose.
4. none of the above.
19.
Which of the following is a purine?
1. adenine.
2. cytosine.
3. thymine.
4. adenoid.
20.
According to "Chargaff's rules" regarding base-pairing,
1. purines = pyrimidines
2. there may be more "A" than "G" in DNA
3. the percent "G + C" varies from organism to organism.
4. all of the above.
21.
If a sample of DNA has a base composition of 20 percent guanine, which of the
following statements is true?
1. the AT to GC ratio must be equal to 1.
2. the ratio of purines to pyrimidines must be greater than 1.
3. the amount of the other purine adenine must also equal 20 percent.
4. the amount of thymine must equal 30 percent.
22.
Each of the following is part of the original Watson-Crick model for DNA EXCEPT
1. DNA consists of antiparallel strands of nucleotides.
2. the DNA helix is right-handed.
3. DNA can be triple-stranded.
4. DNA strands are complementary to one another.
23.
Which of the following is a pyrimidine?
1. adenine.
2. cytosine.
3. guanine.
4. tyrosine.
24.
If one strand of a short DNA fragment has the sequence
5'-GCGCGCGCGC-3' then the other strand of DNA has the sequence
1. 5'-ATATATATAT-3'
2. 5'-GCGCGCGCGC-3'
3. 5'-TATATATATA-3'
4. 5'-CGCGCGCGCG-3'
25. In DNA, a nitrogenous base is attached to the deoxyribose by
1. a phosphodiester bond.
2. hydrogen bonds.
3. an ionic bond.
4. none of the above.
26.
Reassociation kinetics experiments reveal that
1. DNA is very easy to melt.
2. DNA cannot be reannealed once melted.
3. the kinetics of reassociation says nothing about genomic structure.
4. eukaryotic genomes contain many repeated sequences.
27.
Bacterial strain "A" has a genome with a Cot0.5 value of "Y" while bacterial strain "B" has a
genome with a Cot0.5 value of "3Y". If the size of the genome for strain A is "X" base pairs, then the size of
the genome of strain "B" is best approximated by:
1. 3X base pairs.
2. 2X base pairs.
3. X/2 base pairs.
4. X3 base pairs.
28.
In the Meselson-Stahl experiment in which replication in bacteria was shown to be
semiconservative, what was the density distribution of the isolated DNA molecules two generations after
shifting bacteria from "heavy" to "light" growth medium?
1. 100% of the molecules were of heavy density.
2. 100% were of intermediate density
3. 50% were of heavy density, 50% were intermediate density
4. 50% were of light density, 50% were intermediate density.
29. During replication of DNA, strand elongation proceeds
1. in a 5' to 3' direction on the leading strand, but in a 3' to 5' direction on the lagging strand.
2. in a 3' to 5' direction on the leading strand, but in a 5' to 3' direction on the lagging strand.
3. in a 5' to 3' direction on both the leading and lagging strands.
4. in a 3' to 5' direction on both the leading and lagging strands.
30.
Which of the following is required for replication of DNA in bacteria?
1. RNA primase.
2. reverse transcriptase.
3. telomerase
4. all of the above.
31.
In Luria and Delbruck's classic experiments to distinguish between "spontaneous" versus
"adaptive" mutation,
1. evidence for spontaneous mutation was lacking.
2. evidence for adaptive mutation was produced.
3. evidence that DNA is the genetic material was produced.
4. none of the above.
32.
In a Drosophila cross involving an "attached X" female,
1. all male offspring die.
2. the mother passes a Y chromosome to her male offspring only.
3. the mother does not have a Y chromosome.
4. some living female offspring receive only two Y chromosomes.
33.
The "abnormal" tautomeric form of guanine hydrogen bonds well with
1. the abnormal form of adenine.
2. the normal form of guanine.
3. the normal form of cytosine.
4. the normal form of thymine.
34. Intercalating agents
1. usually cause transitions.
2. usually cause chromosomal inversions.
3. may cause slippage during DNA replication, leading to insertions or deletions.
4. should be tried for treason.
35.
A liver extract is often used in the Ames test
1. to inactivate a potential mutagen.
2. because some compounds are normally mutagenic to humans only after being metabolized in
the body.
3. as food for the rats used in the analysis.
4. because lobster extract is too expensive.
36. You finally graduated from college and you take a high-paying job as a genetic councilor. An
individual tells you that a rare disease is being passed along in her family and that the age of onset of this
disease unfortunately appears to be getting earlier and earlier with each successive generation. You
correctly tell this individual that the cause of this disease is likely a
1. transversion.
2. centric fusion.
3. dynamic mutation.
4. recessive allele.
37.
What is the most common type of DNA damage caused by ultraviolet light?
1. DNA breaks.
2. thymine dimers.
3. alkylated bases.
4. apurinic sites.
38. Which statement is true?
1. nucleotide excision repair is defective in all cases of Xeroderma Pigmentosum.
2. it has been firmly established that photoreactivation is defective in certain humans that are
prone to cancer.
3. in humans, strand discrimination in mismatch repair has been shown to be accomplished by
methylation.
4. some cases of Xeroderma Pigmentosum involve a defect in a type of DNA polymerase.
39.
Regarding the SOS response in bacteria:
1. it induces death.
2. if it is defective, it could contribute to the development of cancer.
3. it allows bacteria to survive damage by ultraviolet light but the cost is an increased mutation
rate.
4. it prevents mutations that might otherwise be induced by ultraviolet light.
40. Which statement about DNA repair is true?
1. photoreactivation can occur in the dark.
2. excision repair does not occur in humans.
3. defects in DNA repair may predispose an individual to cancer
4. DNA repair can correct a mutation after the mutation occurs.
THAT'S ALL!