Download Problem Set 1A Due August 31 1. A diploid somatic cell from a rat

Problem Set 1A
Due August 31
1. A diploid somatic cell from a rat has a total of 42 chromosomes (2n = 42). As
in humans, sex chromosomes determine sex: XX in females and XY in males.
i.
What is the total number of DNA molecules in a rat cell in G2? 84
ii. What is the total number of telomeres in a rat cell in G2? 168
iii. What is the total number of chromosomes present in the cell during
metaphase I of meiosis? 42
iv. What is the total number of chromosomes in a polar body cell from a rat?
21
2. A chromosome with a centromere at the very end is called telocentric.
3. Use the following choices for this questions:
a. Meiosis I prophase
b. Meiosis I anaphase
c. Meiosis II prophase
d. Meiosis II anaphase
e. Mitosis prophase
f. Mitosis anaphase
i.
Chromosomes are in unseparated, sister-chromatid form, at the end of the
phase(s) a, b, c, e.
ii. The first stage after which a dividing cell that started as a diploid would be
haploid b.
iii. Sister chromatids separate during d, f.
iv. Chromosomes are randomly partitioned during b, contributing to genetic
diversity.
v. Crossing over (genetic recombination) occurs in a.
4. Write a calculation, with an explanation, for the number of different
combinations of chromosomes in the gametes of a rat (2n = 42).
2n = 221 where n = number of homologous pairs, of which there are 21 in
a (female) rat.
Chapter 2
5. Microscopy to look at a cell's chromosomes is often done when the cell is in
mitotic metaphase. For example, karyotypes that extract chromosomes from
a single cell and photograph them to look for abnormalities are done on
metaphase, rather than interphase, cells. Why?
In metaphase, chromosomes are condensed and are more easily
visualized.
6. The cells illustrated below belong to a species with a diploid chromosome
number of four. Each of the cells below is in which stage of mitosis or
meiosis?
a. meiosis I anaphase
b. mitosis metaphase
c. mitosis anaphase
d. meiosis I anaphase
e. meiosis II metaphase
Chromosomes and Cellular Reproduction
7. Find and describe at least four errors in the drawing below of mitosis
anaphase.
a. Chromosomes that are separating are still duplicated.
b. Spindles are not coming from a common spindle pole body.
c. Sister chromatids do not have identical alleles for the B gene.
d. Two alleles of the D gene are on one chromosome.
e. No alleles of the A gene are on the homologous chromosome.
f. Homologous chromosomes appear to have paired and to be
segregating.
A-
-A
B-
-b
d-
B-
dD-
D-
-b
Chapter 2
8. Two gene loci, A and B, are unlinked (and thus assort independently), and
alleles A and B are dominant over alleles a and b. Indicate the probabilities of
producing the following.
a. An AB gamete from an AaBb individual?
a. 1/4
b. An AB gamete from an AABb individual?
b. 1/2
c. An AABB zygote from a cross AaBb × AaBb?
c. 1/16
d. An AaBb zygote from a cross AaBb × AABB?
d. 1/4
e. An Aabb zygote from a cross AaBb × AAbb?
e. 1/4
f. An AB phenotype from a cross AaBb × AaBb?
f. 9/16
g. An AB phenotype from a cross aabb × AABB?
g. 1 (100%)
h. An aB phenotype from a cross AaBb × AaBB?
h. 1/4
9. While doing field work in Madagascar, you discover a new dragonfly species
that has either red (R) or clear (r) wings. Initial crosses indicate that R is
dominant to r. You perform three crosses using three different sets of redwinged parents with unknown genotype and observe the following data:
Cross
1
2
3
Phenotypes
72 red-winged, 24 clear-winged
4 red-winged
96 red-winged
a. What is the most likely genotype for each pair of parents?
Cross 1 results in a 3:1 ratio of red-winged to clear-winged progeny,
therefore the parents are most likely both Rr. Crosses 2 and 3 result
in only red-winged progeny, therefore the parents are most likely all
RR .
b. Do you think there are a sufficient number of progeny to support each
of your answers in the previous question?
Crosses 1 and 3 have a sufficient number of progeny, but the low
number of progeny from cross 2 precludes making any conclusions.
10. In snapdragons, the allele for red flowers is incompletely dominant over the
allele for white flowers, and thus heterozygotes have pink flowers. What
ratios of snapdragon flower colors would you expect to see among progeny
generated from the following crosses?
a. red × white
100% pink
b. red × pink
50% pink; 50% red
c. white × pink
50% pink; 50% white
e. white × white
100% white
f. pink × pink
25% red; 50% pink; 25% white
g. red × red
100% red
Chromosomes and Cellular Reproduction
11. The red kernel color in wheat is caused by the presence of at least one
dominant allele from each of two independently segregating gene pairs (e.g.,
R-B-). Wheat plants with rrbb genotypes have white kernels, and plants with
genotypes R-bb and rrB- have yellow kernels. You cross a plant true
breeding for red kernels with a plant true breeding for white kernels.
a. What is the expected phenotype(s) and ratios of the F1 plants?
Using P for the leaf color locus and S for the stem type locus: PpSs
× ppSs.
b. What are the relative proportions of the phenotypic classes expected in
the F2 progeny after selfing the F1 progeny?
A possible hypothesis is that each trait is controlled by
independently assorting single gene pairs, and that one allele in
each pair exhibits complete dominance over the other. Class 1 = 3/8,
class 2 = 3/8, class 3 = 1/8, class 4 = 1/8.
12. List at least four phenomena that can alter expected Mendelian phenotypic
ratios in genetic crosses.
(1) Linkage
(2) Epistasis
(3) X-linked genes
(4) Lethal recessive alleles
(5) Environmental effects (sex-influenced traits, etc.)
(6) Continuous traits
(7) Variable expressivity
13. Albinism is a somatic recessive condition resulting from the inability to
produce the dark pigment melanin in skin and hair. A man and woman with
normal skin pigmentation have two children. The man has one albino parent;
the woman has parents with normal pigmentation, but an albino brother.
a. What is the probability that at least one of the children is albino?
Using alleles A and a, since one of the man’s parents is albino, he must
be a carrier (Aa) of the recessive allele (probability = 1). The woman has
an albino brother, which means both her parents must be carriers (Aa).
However, the woman (who is not albino) could have either an AA or Aa
genotype. In the woman’s case the aa (albino) genotype must be
excluded as a possibility, therefore the final probability of the woman
having either AA or Aa genotypes is 1/3 and 2/3, respectively.
Case 1: If the woman is Aa, the probability of at least one of the two
children being albino from the cross (Aa × Aa) equals the probability
that one child is albino (1/4) × the probability that the other child is
normal (3/4) × 2 (must multiply by 2 because either birth order is
possible for this outcome), plus the probability that both children are
Chapter 2
albino, (1/4) × (1/4) = 1/16. The probability that at least one child is
albino equals 2(3/4)( 1/4) + (1/16) = 6/16 + 1/16 = 7/16 (43.75%). Because
the probability of the woman having the Aa genotype is 2/3, the final
probability of her having albino children is (7/16)(2/3) = 29.2%.
Case 2: If the woman is AA, the probability of at least one of the two
children being albino from the cross (Aa × AA) is zero.
b. What is the probability of both children being albino?
As noted above, if the woman is Aa, then the probability of both
children being albino (aa) equals (1/4) × (1/4) = 1/16 (6.25%). Because
the probability of the woman having the Aa genotype is 2/3, the final
probability of both her children being albino is (1/16)(2/3) = 4.2%.
14. In order to put yourself through school, you decide to start commercially
breeding parrots. In one of your lines you discover two birds (a male and a
female) with brilliant purple feathers. You cross these parents several times
and observe 20 progeny with brilliant purple feathers and 10 with wild-type
color. Offer an explanation for this ratio.
If the allele for purple feathers (P) is dominant to wild-type color (p) and
the parental birds are heterozygous you would expect to see a 3:1 ratio
of purple (PP and Pp) and wild-type (pp) progeny. If the P allele also
acts as a recessive lethal (i.e. lethal in homozygous state) then the PP
class would never occur in the progeny resulting in a 2:1 purple:wildtype ratio. The observed progeny ratio is consistent with 2:1
segregation.