1. (2 pts.) Sister chromatids in a replicated chromosome that is ready to enter mitosis are held together at a specific region of the chromosome called the
A. Centromere
B. Mitotic spindle
C. Centrosome
D. Chiasma
Answers seen: A (60), B (0), C (3), D (0)
2. (2 pts.) The mode of cell division that results in four daughter cells that are genetically different, with each cell containing one haploid set of chromosomes, is known as
A. Mitosis
B. Cell Division
C. Meiosis
D. Interphase
E. Diakinesis
Answers seen: A (0), B (0), C (63), D (0), E (0)
3. (2 pts.) Which of the following results from a physical exchange between chromatids of homologous chromosomes in meiosis?
A. Bivalent
B. Meiotic division
C. Chiasma
D. Tetrad
E. Synapsis
Answers seen: A (9), B (4), C(50), D(1), E (4)
4. (2 pts.) Starting with a cross between AA and aa, and following that with an F1 intercross, will produce what proportion of heterozygotes in the F2?
A. 1/8
B. 1/4
C. 1/3
D. 1/2
E. All heterozygotes
Answers seen: A (0), B (2), C (0), D (56), E (6)
5. (2 pts.) The probability that each of four children in a family will be male is
A. 1/16 (0.062)
B. 1/4 (0.25)
C. 1/2 (0.50)
D. 1/64 (0.016)
E. 1/8 (0.125)
Answers seen: A (24), B (3), C (33), D (0), E (3)
6. (2 pts.) In dihybrid crosses, the ratio 9:3:3:1 in the F2 indicates
A. Codominance
B. Independent assortment
C. Incomplete dominance
D. Linkage
Answers seen: A (6), B (53), C (3), D (2)
7. (2 pts.) Mating of two organisms produces a 1:1 ratio of phenotypes in the progeny. The parental genotypes are
A. Aa x Aa
B. Aa x aa
C. AA x aa
D. AA x AA
E. There is not enough information to solve the problem.
Answers seen: A (2), B (55), C (5), D (1), E (1)
8. (2 pts.) The presence in a population of more than two alleles of a gene is called
A. Mutant
B. Codominant
C. Pleiotropic
D. Multiple alleles
E. Dihybrid
Answers seen: A (1), B (4), C (22), D (37), E (1)
9. (2 pts.) The maximum observable frequency of recombination between two genes is
A. 100%
B. 80%
C. 50%
D. 10%
E. 1%
Answers seen: A (4), B (4), C (55), D (1), E (0)
Explanation: Please see this.
10. (2 pts.) The coupling relationship or phase (coupling or repulsion) of alleles in a doubly heterozygous parent is detectable as the
A. Most frequent type of gametes
B. Lowest incidence of crossing over
C. Largest number of map units between corresponding alleles
D. Highest frequency of recombination
Answers seen: A (20), B (16), C (7), D (21)
Explanation: Please see this.
11. (5 pts.) Three genes (A, B, and C) show independent assortment. What is the probability that AaBbCc x AAbbCc will produce Aabbcc?
A. 1/2
B. 1/4
C. 1/8
D. 1/16
E. 1/32
Answers seen: A (0), B (2), C (8), D (47), E (7)
Explanation: The probability of Aa from Aa x Aa is 1/2. The probability of bb from Bb x bb is 1/2. The probability of cc from Cc x Cc is 1/4. The probability that AaBbCc x AAbbCc will produce Aabbcc is (1/2)(1/2)(1/4) = 1/16.
Please see this.
12. (5 pts.) A man known to have a normal karyotype (46, XY) is married to a woman known to have a normal karyotype (46, XX). They have a son who is XYY (47, XYY). The best explanation for this is
A. Nondisjunction at meiosis I in the father.
B. Nondisjunction at meiosis II in the father.
C. Nondisjunction at meiosis I in the mother.
D. Nondisjunction at meiosis II in the mother.
E. Nondisjunction at meiosis I in both parents.
Answers seen: A (13), B (49), C (0), D (1), E (0)
Explanation: The only possible origin of an XYY male from karyotypically normal parents is the fertilization of an X-bearing egg by a YY-bearing sperm. This means that nondisjunction occurred in the father. Because the two sex chromosomes in the disomic sperm are sisters, rather than homologs, the disomic gamete resulted from the failure of the equational division, meiosis II.
13. In Drosophila, black body is caused by a recessive mutation symbolized b, while cinnabar eyes are caused by a recessive mutation symbolized cn. Females that are phenotypically wild type (tan body, brick-red eyes) are crossed to homozygous black cinnabar (b cn / b cn) tester males. Four progeny types are seen in the frequencies shown below.
| Phenotype | Genotype | Number |
| wild type | + + / b cn | 27 |
| black cinnabar | b cn / b cn | 23 |
| black | b + / b cn | 239 |
| cinnabar | + cn / b cn | 211 |
a. (5 pts.) What is the genotype of the female parents?
A. + + / + +
B. b cn / + +
C. b + / + cn
D. b b / cn cn
E. There is not enough information to solve the problem.
Answers seen: A (5), B (14), C (45), D (0), E (0)
Explanation: The most frequent gamete types are b + and + cn, so these are the parental types.
Please see this.
b. (5 pts.) What is the frequency of recombination between b and cn?
A. 2%
B. 10%
C. 23%
D. 27%
E. 50%
Answers seen: A (1), B (41), C (5), D (2), E (15)
Explanation: 27 + 23 = 50 recombinants, 239 + 211 = 450 nonrecombinants, the total is 500. 50/500 = 10%.
Please see this.
14. (10 pts.) Four genes are linked. In a series of testcross experiments, the frequency of recombination between a and b is 19%; between a and c is 48%; between a and d is 10%; between b and d is 10%; between c and d is 39%. What is the order of these four genes?
A. a – d – b – c
B. b – a – c – d
C. b – c – d – a
D. b – a – c – d
E. a – b – c – d
Answers seen: A (46), B (3), C (0), D (3), E (2)
Explanation: Let's show the data this way:
| gene pair | recombination |
| a - d | 10% |
| b - d | 10% |
| a - b | 19% |
| c - d | 39% |
| a - c | 48% |
The two pairs of genes with the smallest fraction of recombinants are a - d and b - d. The d gene is common to these pairs. There are two possible orders of these three genes. Either the order is a - d - b, in which case we would expect the a - b distance to be around 20%, or the order is d - a - b or d - b - a with 0% between a and b in either case. We observe an a - b distance of 19%, so the correct order is a - d - b.
Now we need to place c. The c gene is closer to d (39%) than it is to a (48%), so the order is a - d - b - c.
15. Three independently assorting genes control coat color in the mouse: brown (b), which is recessive to the wild type allele B; dilute (d), which is recessive to the wild type allele D; and colorless (c), which is recessive to the wild type allele C. Mice homozygous for colorless (cc) are pink-eyed albinos; colorless is epistatic to brown and dilute. There are five possible phenotypes: black ( B- C- D- ), albino (cc), brown ( bb C- D- ), dilute ( B- C- dd ), and brown dilute ( bb C- dd ).
a. (5 pts.) An albino mouse is testcrossed to a true-breeding brown dilute stock ( bb CC dd ). The mating produces a set of litters with a total of 15 dilute and 11 black mice. The genotype of the albino mouse is
A. BB cc DD
B. Bb cc DD
C. Bb cc Dd
D. BB cc Dd
E. bb Cc dd
Answers seen: A (0), B (1), C (11), D (52), E (0)
Explanation: Please see this.
b. (5 pts.) An albino mouse is testcrossed to a true-breeding brown dilute stock ( bb CC dd ). The mating produces a set of litters with a total of 9 brown dilute, 7 dilute, 7 brown, and 8 black mice. The genotype of the albino mouse is
A. BB cc DD
B. Bb cc DD
C. Bb cc Dd
D. BB cc Dd
E. bb Cc dd
Answers seen: A (2), B (2), C (57), D (2), E (1)
Explanation: Please see this.
16. The pedigrees below show three families with the ABO blood types of each individual indicated.
a. (2 pts.) Could individual 2 be the father of individual 3? ____No____ (Yes or No)
b. (2 pts.) Could individual 2 be the father of individual 4? ____Yes___ (Yes or No)
c. (2 pts.) Could individual 6 be the father of individual 7? ____Yes____ (Yes or No)
d. (2 pts.) Could individual 6 be the father of individual 8? ____Yes____ (Yes or No)
e. (2 pts.) Could individual 10 be the father of individual 11? ___Yes_____ (Yes or No)
Answers seen: a Y(1) N(63); b Y(64) N(0); c Y(62) N(2); d Y(61) N(3); e Y(52) N(13)
Explanation: Please see this.
17. (5 pts.) A rare human disease affected a family as shown in the pedigree below.
The most likely mode of inheritance is
A. Autosomal dominant
B. Autosomal recessive
C. X-linked dominant
D. X-linked recessive
E. There is not enough information to solve the problem.
Answers seen: A (3), B (53), C (2), D (5), E (0)
Explanation: We first ask ourselves whether there are any affected individuals who have two unaffected parents. The answer is yes, so we are dealing with a recessive allele. We then ask if there are any females with the recessive phenotype who have a father or a son with the dominant phenotype. The female in the second generation has an unaffected father, so the recessive allele here cannot be sex-linked. This is an autosomal recessive.
Please see this.
18. (5 pts.) A rare human disease affected a family as shown in the pedigree below.
The most likely mode of inheritance is
A. Autosomal dominant
B. Autosomal recessive
C. X-linked dominant
D. X-linked recessive
E. There is not enough information to solve the problem.
Answers seen: A (12), B (5), C (38), D (4), E (1)
Explanation: We first ask ourselves whether there are any affected individuals who have two unaffected parents. The answer is no, so we are dealing with a dominant allele. There are no unaffected females who have a father with the dominant allele. We see a total of seven affected daughters whose fathers are affected, and no unaffected daughters from these fathers. If this is an X-linked dominant, we would expect those results 100% of the time. If this is an autosomal dominant, we would expect seven affected and zero unaffected daughters only (1/2)7 or in 1/128 of such samples.
It is also informative to notice that affected fathers do not transmit the disease to their sons. There are five sons of affected fathers, all unaffected. While it is possible that the disease allele is autosomal dominant, a better hypothesis is that the disease allele is an X-linked dominant. We expect the five unaffected sons of affected fathers 100% of the time under the hypothesis of X-linked dominant, while we expect the five unaffected sons of affected fathers only (1/2)5 or in 1/32 of such samples.
Taken together, the evidence strongly supports the model of an X-linked dominant. We would expect these results from an autosomal dominant only (1/128)(1/32) or in 1/4096 of such samples.
Please see this.
19. (5 pts.) Red-green colorblindness is an X-linked trait in humans. A man with normal color vision is married to a woman with normal vision whose father was colorblind. They have a daughter with Turner Syndrome (45, X0) who is colorblind. The best explanation for this is
A. Nondisjunction at meiosis I in the father
B. Nondisjunction at meiosis II in the father
C. Nondisjunction at an unknown division in the father
D. Nondisjunction at meiosis I in the mother
E. Nondisjunction at meiosis II in the mother
Answers seen: A (5), B (12), C (38), D (7), E (1)
Explanation: Using the symbol c to represent colorblindness, a man with normal color vision is C/Y. A woman whose father was colorblind is C/c. Their daughter with Turner Syndrome is colorblind, so she inherited a single X chromosome from her mother. We know that her single X chromosome is maternally derived because she is colorblind. Normal eggs have a single X chromosome, so meiosis in the mother was normal. Her father failed to contribute an X chromosome; his daughter resulted from a nullo-X, nullo-Y sperm. Because there aren't any sex chromosomes in the sperm, we know that it resulted from a nondisjunction event, but we can't tell whether it occurred at the first or the second division.
20. True-breeding Siamese cats with color at the "points" are homozygous for a temperature-sensitive allele of the tyrosinase gene, cscs. There are also true-breeding albino cats, cc. True-breeding black cats are CC. In this series of alleles, C > cs > c, so C- are black and csc are Siamese.
a. (5 pts.) A cat with color at the points is mated to a black cat, producing a litter of three kittens: one black, one with color at the points, and one albino. The genotypes of the parents in this cross are
A. cscs x Cc
B. csc x Cc
C. csc x Ccs
D. cscs x Ccs
E. There is not enough information to solve the problem.
Answers seen: A (1), B (56), C (3), D (1), E (1)
Explanation: A cat with color at the points is cscs or csc. A black cat can be CC, Ccs, or Cc. If any albino (cc) kittens are seen, the cat with color at the points is csc and the black cat is Cc. The expected ratio is 1:2:1, but with a sample size of three kittens, we shouldn't take this very seriously.
b. (5 pts.) Two cats with color at the points are mated, producing a litter of four kittens: two with color at the points and two albinos. The genotypes of the parents in this cross are
A. cscs x cscs
B. cscs x csc
C. csc x csc
D. csc x cc
E. There is not enough information to solve the problem.
Answers seen: A (0), B (2), C (51), D (7), E (2)
Explanation: A cat with color at the points is cscs or csc. If any albino (cc) kittens are seen in the cross of two cats with color at the points, both of the parents had to be csc. The expected ratio is 3:1, but with a sample size of four kittens, we shouldn't take this very seriously.
c. (5 pts.) A cat with color at the points is mated to an albino, producing a litter of four kittens: two with color at the points and two albinos. The genotypes of the parents in this cross are
A. cscs x csc
B. csc x csc
C. csc x cc
D. cscs x cc
E. There is not enough information to solve the problem.
Answers seen: A (0), B (1), C (55), D (4), E (1)
Explanation: A cat with color at the points is cscs or csc. If any albino (cc) kittens are seen in the cross to albino (cc), the cat with color at the points had to be csc. The expected ratio is 1:1, but with a sample size of four kittens, we shouldn't take this very seriously.