L111 Exam II, FRIDAY, October 14, Fall Semester of 2005
NAME __________________________________________
ID NUMBER ___________________
Please READ CAREFULLY and FOLLOW these directions:
1] Use a No. 2 lead pencil ONLY
2] Fill out the enclosed SCANTRON SHEET
LAST NAME FIRST
a) NAME---b) Do NOT WRITE in the SPECIAL CODES area
3] SIGN THE ANSWER SHEET ACROSS THE TOP MARGIN (above the NAME heading). We
will NOT accept any unsigned answer sheets. By signing the answer sheet, you are stating that you have
neither given nor received answers or information from another student during the examination.” By
signing, you are ALSO confirming that you understand that, if you violate this policy, you will receive a
score of “0” for this examination and, depending upon the severity of the offense, may receive an “F” for
the L111 course and a letter describing your offense placed in your permanent file in the Dean’s office.
4] READ EACH QUESTION carefully and FILL IN THE CIRCLE that corresponds to the SINGLE
BEST ANSWER. NO PARTIAL CREDIT WILL BE GIVEN FOR ANSWERS THAT ARE CORRECT
BUT NOT THE BEST ANSWER.
5] The exam consists of 27 questions each question is worth 4 points.
6] Questions number 26 and 27 are BONUS questions. If you answer them correctly, each is worth the
equivalent of 1 regular question. If you answer them incorrectly, they will not be counted at all.
7] Exams must be turned in by 9:55 AM
8] When you have completed the exam, TURN IN YOUR SIGNED SCANTRON SHEET to the AI or UTI
at the front of the room and SHOW YOUR STUDENT ID CARD.
9]SAVE THE QUESTIONS. You will need them in your Discussion Sections this coming week, where
you will re-take the test after talking about the questions with your group.
1. In a randomly mating, diploid Mendelian population, in the absence of evolutionary
forces, the Genotype Frequency Distribution
a.
b.
c.
d.
Changes at each generation.
Reaches a stable equilibrium after many generations.
Remains constant over time.
Reaches a stable equilibrium after a single generation.
e. (c) and (d) are True.
.
2. Imagine a gene with four alleles, A1 A2, A3, and A4 in a very large population. Which
population has the most possible genotypes?
a. A population of haploids.
b. A population of diploids.
c. A population of triploids.
d. A population of tetraploids.
e. All of the above have an equal number of possible genotypes.
3. If a population consists of individuals of THREE genotypes, BB, Bb, and bb, and the
genotype frequencies are GBB = 0.70, GBb = 0.20, and Gbb = 0.10, then the frequency
of the B allele, PB, is
a. {0.70 + (1/2)(0.20)}.
b.
c.
d.
e.
{0.70 + 0.10}/2
√(0.70)
{(2)(0.70) + (1)(0.70)}
{0.70 + 0.20}.
4. If alleles at gene A affect mouse coat color and alleles at gene B affect mouse hair
length, then genes A and B are linked if
a. Both genes are polymorphic.
b. The traits violate Mendel’s Law of Independent Assortment.
c. The mice mate at random.
d. All of the above.
e. None of the above.
5. An evolutionary consequence of pleiotropy is
a. Selection on one trait results in change in another trait.
b.
c.
d.
e.
The Hardy-Weinberg Equilibrium.
Mutation-selection balance.
A violation of Mendel’s Law of Dominance.
All of the above.
6. The Map from Phenotype to Genotype is complex because of
a.
b.
c.
d.
Epistasis.
Environmental effects.
Predation.
Polypoidy.
e. (a) and (b).
7. Which of the following can cause a phenotypic resemblance among siblings
a. Similar environments during development.
b. Similar maternal effects during development.
c. Similar genetic composition.
d. All of the above.
e. None of the above.
8. Imagine a randomly mating population consisting of N individuals with no
evolutionary forces acting. If the frequency of bb homozygotes is 0.81, then the
frequency of the b allele is
a. 2(0.81)(N).
b. 1 – 0.81.
c. The square root of 0.81.
d. 0.81 + (1/2)(0.09)
e. (0.81)2.
9. An individual with two copies of the same allele is
a. Heterozygous.
b. Polyploid.
c. Parthenogenic.
d. Homozygous.
e. (A) and (C).
10. Paternal Half-sibs share what fraction of their alleles at a locus
a. 0.00.
b. 0.25.
c. 0.50.
d. 0.75.
e. 1.00.
11. It there was a gene with 3 alleles, how many diploid genotypes could exist?
a.
b.
c.
d.
2.
3.
4.
5.
e. 6.
12. In the absence of any evolutionary forces, the genotype frequency distribution in the
fertilized eggs of a species, like the blue fin tuna, that reproduces by Broadcast
Scattering
a. Sums to less than 1.0 because some eggs are not fertilized.
b. Is in the Hardy-Weinberg Equilibrium.
c. Is unpredictable because the gametes are scattered about the ocean.
d. Deviates from Mendel’s Laws.
e. Has no heterozygotes.
13. What effect does reproduction have on genetic variation in a population with
Mendelian inheritance?
a.
b.
c.
d.
Genetic variation is reduced by 1/2 at each generation.
Genetic variation is reduced by 1/3 at each generation.
Genetic variation is increased by 1/2 at each generation.
Dominant alleles increase in frequency at each generation.
e.
Genetic variation is unchanged.
14. Deviations away from the Hardy-Weinberg Equilibrium (HWE) can be caused by
a.
b.
c.
d.
e.
Random mating.
Nonrandom mating.
Natural Selection.
Pleiotropy.
(a) and (c).
15. TRUE or FALSE: If you know the allele frequency distribution, you can always
calculate the genotype frequency distribution.
a. TRUE.
b. FALSE.
16. In a population, positive assortative mating, where ‘like mates with like,’ results in
a. More homozygous offspring than random mating.
b.
c.
d.
e.
More heterozygous offspring than random mating.
More polymorphic offspring than random mating.
More offspring than random mating.
Offspring with greater fitness than random mating.
17. In a randomly mating population, with parental genotype frequencies {GYY, GYy,
Gyy} = {0.25, 0.70, 0.05}, the frequency of matings between YY males and yy
females is
a. 0.25 + 0.05.
b. 0.25 + (1/2)(0.70).
c. (0.25)2
d. 0.25 x 0.05
e. None of the above.
18. The larvae of the gall fly, Eurostra solidaginis, experience
a. Directional selection for smaller size as a result of wasp predation on galls.
b. Stabilizing selection for intermediate size.
c. Disruptive selection as a result of predation by wasps and birds.
d. Directional selection for larger size as a result of bird predation on galls.
e. All of the above.
19. Mutation is a very weak evolutionary force, while Natural Selection against a lethal
recessive allele can be a very strong force. A population achieves mutation-selection
balance when
a. The frequency of the recessive allele is equal to the frequency of the dominant
allele.
b. The deleterious recessive allele is driven out of the population and its
frequency is zero.
c. When heterozygotes have higher relative fitness than either homozygote.
d. The frequency of the recessive allele equals the square root of
the mutation rate.
e. When the frequency of affected individuals is equals the square of the
frequency of the recessive allele.
20. The rate at which natural selection removes a recessive lethal allele from a population
a. Decreases as the allele becomes rarer.
b. Increases as the allele becomes rarer.
c. Remains constant because of the Hardy-Weinberg Equilibrium.
d. All of the above.
e. (a) and (b).
21. The Figure above shows
a. The survival of gall fly larvae as a result of bird predation.
b. The survival of gall fly larvae as a result of wasp predation.
c. Directional selection for smaller gall size.
d. Pleiotropy for body size.
e. Intermediate size galls have higher fitness than large or small galls.
22. TRUE or FALSE. The frequency of the recessive ‘s’ allele is greater than the
frequency of individuals affected with Sickle Cell Disease in both the US and West
Africa.
a. True.
b. False.
23. Relative fitness is
a. The difference between the number of offspring born and the number dying.
b. The size of a gall relative to the size of the predators on the gall.
c. The speed of a zebra relative to the speed of the lion, its major predator.
d. The fitness of a genotype relative to the population’s average
absolute fitness.
e. All of the above are definitions of relative fitness.
24. Consider a population with 1,000 individuals consisting of genotype frequencies
{GYY, GYy, Gyy} = {0.74, 0.24, 0.02} and where yy individuals suffer a lethal genetic
disease. How many copies of the y allele are exposed to natural selection?
a. (2)(1000)(0.02 + [1/2][0.24]).
b. (2)(1000)(√0.02).
c. (2)(1000)(0.02).
d. (1)(1000)(0.24).
e. (1)(0.74 + 0.24 + 0.02) = 1.
25. Which population will experience the strongest Random Genetic Drift?
a.
b.
c.
d.
A population with 250 breeding males and 250 breeding females.
A population with 50 breeding males and 50 breeding females.
A population with 25 breeding males and 25 breeding females.
A population with 500 breeding males and 500 breeding females.
e. A population with 10 breeding males and 10 breeding females.
BONUS QUESTIONS: EACH IS WORTH 4 POINTS.
26. In a randomly mating population, with parental genotype frequencies {GYY, GYy,
Gyy} = {0.16, 0.48, 0.36} and relative fitnesses {wYY, wYy, wyy} = {0.25, 1.05, 0.80}
a. There is heterozygote advantage.
b. The frequency of yy homozygotes after selection equals 0.36 x 0.80.
c. The frequency of the Y allele is less than the frequency of the y allele.
d. All of the above are True.
e. None of the above are True.
27. Random Genetic Drift and Migration have opposite effects on which of the
following
a. The genetic variation within populations.
b. The genetic variation between populations.
c. The total genetic variation.
d. (a) and (b).
e. (a), (b) and (c).