Download Problem set 3 with answers

BIO322: Genetics
Douglas J. Burks
Department of Biology
Wilmington College of Ohio
Problem Set 3
Due @ 10:35 AM January27, 2011
Chapter 4: Problems 3, 5, 12, 23, 25, 31, 37, and 41.
Chapter 5: Problems 8, 14, 24, and 33.
Problem #3. The figure that follows shows the metaphase chromosomes of a male of a particular
species. These chromosomes are prepared s they would be for a karyotype, but they have not yet
been ordered in pairs of decreasing size.
a.
b.
c.
d.
e.
How many centromeres are shown? 7
How many chromosomes are shown? 7
How many chromatids are shown? 14
How many homologous chromosomes are shown? 3
How many chromosomes on the figure are metacentric? Acrocentric? 2 metacentric
and 5 acrocentric
f. What is the likely mode of sex determination in this species? What would you predict to
be different about the karyotype of female in this species? This appears to be an XO
system in which males have a single X and no Y and females have XX. I
would predict that females would have two of the small chromosome.
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Problem #5. Indicate which of the cells numbered i – v matches each of following stages of
mitosis:
a.
b.
c.
d.
e.
Anaphase iii
Prophase i
Metaphase iv
G2 ii
telophase/cytokinesis v
Problem #12. The five cells shown in figure a – e below are from the same individual. For each
cell, indicate whether it is in mitosis, meiosis I, or meiosis II. What stage of cell division is
represented in each case? What is n in this organism?
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a.
b.
c.
d.
e.
f.
is anaphase I of meisosis
is metaphase of mitosis
is Telophase II of meiosis
is anaphase of mitosis
is metaphase II of meiosis
1n = 3
Problem #23. A system of sex determination known as haplodiploidy is found in honeybees.
Females are diploid, all males (drones) are haploid. Male offspring result from the development
of unfertilized eggs. Sperm are produced by mitosis in males and fertilize eggs in the females.
Ivory eye is a recessive characteristic in honeybees; wild-type eyes are brown.
a. What progeny would result from an ivory-eyed queen and a brown-eyed drone? Give
both genotype and phenotype for progeny produced from fertilized and nonfertilized
eggs. ♀ ii and ♂ I are P generation. F1 = ♀ iI brown and ♂ i ivory.
b. What would result from crossing a daughter from the mating in part a with a browneyed drone? P = ♀ iI and ♂I F1 ♀ iI and II both brown and ♂I and I 50%
ivory and 50% brown.
Problem #25. Barred feather pattern is a Z-linked dominant trait in chickens. What offspring
would you expect from
(a) the cross of a barred hen to a non barred rooster
Males are WW and females are ZW. The female is Z/WB and the male is
Wb/Wb
F1 ♂ WBWb, barred ♀ ZWb nonbarred
(b) the cross of an FI rooster from part (a) to one of his sisters? ♂ WBWb, barred x ♀
ZWb nonbarred F2 ♂ WBWb barred and WbWb non-barred ♀ ZWb nonbarred and ZWB barred.
Problem 31. The pedigree that follows indicates the occurrence of albinism in a group of Hopi
Indians, among whom the trait is unusually frequent. Assume that the trait is fully penetrant (all
individuals with a genotype that could give rise to albinism will display this condition).
a. Is albinism in this population caused by a recessive or a dominant allele?
The trait skips generations in several cases. Since the trait is fully
penetrant and common this indicates recessive.
b. Is the gene sex-linked or autosomal?
If the trait is sex linked male II-7 should be deaf as should male as
should male III-5. The trait must be autosomal.
What are the genotypes of the following individuals?
c. individual I-1 dd
d. individual 1-8 Dd
e. individual 1-9 Dd
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f. individual II-6 Dd
g. individual II-8 Dd
h. individual III-4 Dd
Problem #37. Several different antigens can be detected in blood tests. The following four traits
were tested for each individual shown:
(IA and IB codominant, i recessive)
(Rh + dominant to Rh-)
type (M and N codominant)
(Xg(a+) dominant to Xg(a-)
ABO type
Rh type
MN
Xg(a) type
All of these blood type genes are autosomal, except for Xg(a), which is X linked.
Mother
AB
RH- (RH-RH-)
MN
Xg(a+)
Daughter
A
AB
A
B
O no
RH+ (RH+RH-)
MN
Xg(a-) (a-a-)
RH+
M
RH- No
N
Xg(a+) No sex
linked
Xg(a-)
RH+
N
Xg(a-)
RH-
MN
Xg(a-)
Alleged father 1
Alleged father 2
Alleged father 3
Alleged father 4
a. Which, if any, of the alleged fathers could be the real father? Male 3
b. Would your answer to part a change if the daughter had Turner syndrome (the
abnormal phenotype seen in XO individuals)? If so, how? In this case if it was the male
who did not contribute an X then both male 1 or 3 could be father.
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Problem #41. The pedigree at the bottom of the page shows the inheritance of various types of
cancer in a particular family. Molecular analysis (described in subsequent chapters) indicate that
with one exception, the cancers occurring in the patients in this pedigree are associated with a
rare mutation in a gene called BRCA2.
Which individual is the exceptional cancer patient whose disease is not associated with a
BRCA2 mutation?
I would say individual III-5 since he exhibits a cancer but is not
part of the lineages carrying the BRCA2 allele. If we knew the cause
of death of others they too might be exceptions.
Is the BRCA2 mutation dominant or recessive to the normal BRCA2 allele in terms of its
cancer-causing effects?
If you assume that individuals who are deceased may have exhibited
some effects and that expressitivity and penetrance are not complete
then I would say dominant.
Is the BRCA2 gene likely to reside on the X chromosome, the Y chromosome, or an
autosome? How definitive is your assignment of the chromosome carrying BRCA2?
As a hypothesis, I would say sex linked. More females than males
show the trait which is expected and males who show the trait have
daughters with the trait. The pedigree is also consistent with
autosomal. I would need to see several more pedigrees. If autosomal
then the trait is sex-limited to some extent.``
Is the penetrance of the cancer phenotype complete or incomplete?
Individuals II-1 and II-2 indicate that penetrance is incomplete
Is the expressivity of the cancer phenotype unvarying or variable?
Appears variable
Are any of the cancer phenotypes associated with the BRCA2 mutation sex-limited or
sex-influenced?
Males can’t have ovarian cancer and there is no mention of
testicular cancer so this is sex-limited. The breast cancer penetrances
appears to be sex influenced. One could study estrogen and other sex
hormone levels in males who exhibit breast cancer.
How can you explain the absence of individuals diagnosed with cancer in generations I
and II? Low penetrance and/or incomplete records of cause of death. Length of life
different or not good records.
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Problem #8. In Drosophilia, males from a true-breeding stock with raspberry-colored eyes were
mated to females from a true-breeding stock with sable-colored bodies. In the F1 generation, all
the females had wild-type eye and body color, while all the males had wild-type eye color but
sable-colored bodies. When F1 males and females are mated, the F2 generation was composed of
216 females with wild type eyes and bodies, 223 females with wild-type eyes and sable-bodies,
191 males with wild-type eyes and sable-bodies, 188 males with raspberry eyes and wild-type
bodies, 23 males with wild-type eyes and bodies, and 27 males with raspberry eyes and sable
bodies. Explain these results by diagraming crosses, and calculate any relevant map distances.
P = ♂ + raspberry and ♀ sable +. That the males have the mothers phenotype and
females are wild type for both traits suggest sex linkage P = ♂ + ras/Y and ♀
sab sab ++ F1 is ♂ sab+/Y and ♀ sab +/ + ras. Notice the linkage
Let’s look at males from F2 since they inherit there X genes from the mother and
only have one copy so that their phenotype reflects their genotype directly.
♂
sab +
+ ras
++
Sab ras
191
188
23
27
429
Recombinants/ total
0.117
m.u.
m.u
P = Female
Male
F1 = Female
Males
F2 = Females
= males
parental types
Recombinant types
= RF
= Rf
= RF x 100
= 11.7
++, ss
r+, Y
r+, s+
+s, Y Sex linked
?+, +s = 223
?+,ss = 188
+ s , Y = 191
r+, Y =188
++, Y = 23
Rs, Y = 27
Problem # 14. In corn, the allele A allows the deposition of anthocynanin (blue) pigment in the
kernels (seeds), while aa plants have yellow kernels. At a second gene, W produces smooth
kernels, while ww kernels are wrinkled. A plant with blue smooth kernels was crossed to a plant
with yellow wrinkled kernels. The progeny consisted of 1447 blue smooth, 169 blue wrinkled,
186 yellow smooth, and 1510 yellow wrinkled.
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a. Perform a Chi Square analysis of data.
Predict a 1:1:1:1. Total sample is 3312 so you expect 828 of each type.
X2 = [(1447 – 828)2/828] + [(169 – 828)2/828] + [(186– 828)2/828]
+ [(1510– 828)2/828]
2
X = 462 + 524 + 497 + 561 = 2041
actual
1447
169
186
1510
predicted
828
828
828
828
2
X =
[(= a -p)2/p]
462.8
524.5
497.8
561.7
2046.8
p=
0.0000
Null hypothesis (genes are not linked) is rejected. Genes are linked.
b. Are the a and w loci linked? If so how far apart are they?
RF = (169 + 186)/ 3312 = 0.107 m.u. = 0.107 x 10 = 10.7 m.u.
c. What was the genotype of the blue smooth parent? Include the chromosome arrangement
of alleles.
Since all four phenotypes are present then AaWw x aaww
Aw/aW X aw/aw
d. If a plant grown from a blue wrinkled progeny seed is crossed to a plant grown from a
yellow smooth F1 seed, what kinds of kernels would be expected and in what
proportions?
A- w/ a-w x a-W/a-w All four possible phenotypes in approximately 1:1:1:1 since
opposite parents are homozygous recessive for opposite traits.
Problem # 24. In the tubular flowers of foxgloves, wild-type coloration is red while a mutation
called white produces white flowers. Another mutation, called peloria, causes the flowers at the
apex of the stem to be huge. Yet another mutation called dwarf, affects the stem length. You
cross a white-flowered plant (otherwise phenotypically wild type) to a plant that is dwarf and
peloria but has wild type red flower color. All of the F1 plants are tall with white normal-sized
flowers. You cross an F1 plant back to the dwarf and peloria parent, and you see 543 progeny
shown in the chart. (only mutants traits are noted.)
dwarf, peloria
white
dwarf, peloria, white
wild type
dwarf, white
peloria
dwarf
peloria, white
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dp+
++w
dpw
+++
d+w
+p+
d ++
+pw
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172
162
56
48
51
43
6
5
a. Which alleles are dominant?
In the F1 you observe white (dominant ) tall (dominant) and normal
flower size normal (dominant)
b. What were the genotypes of the parents in the original cross?
dp+/++w x ddwwrr
c. Perform a Chi Square analysis to determine if linkage is present
Actual
Predicted
(a-p)2/p
172
67.875
159.735
162
67.875
130.5269
56
67.875
2.077578
48
67.875
5.819751
51
67.875
4.195442
43
67.875
9.116252
6
67.875
56.40539
5
67.875
58.24332
2
X
426.1197
P less than 0.05 genes are linked
d. Draw a map showing the linkage relationships of these three loci?
Parental linkage is and dco are
d++
dp+
++w
+pw
d to p d+ and +p
dwarf, peloria
white
dwarf, peloria, white
wild type
dwarf, white
peloria
dwarf
peloria, white
dp+
++w
dpw
+++
d+w
+p+
d ++
+pw
19.33 d—p
21.17 p – w
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172
162
56
48
51
43
6
5
e. Is there interference? If so, calculate the coefficient of coincidence and the interference
values.
Yes the expected number of dco is 22 and the observed is 11. 11/22 = .5 for co
and interference is 50%
Problem #33. Neurospora of genotype a + c are crossed with Neurospora of genotype + b +.
The following tetrads are obtained (note the genotype of the four spore pairs in an ascus are
listed, rather than listing all eight spores):
a+c
a+c
+b+
+b+
137
abc
abc
+++
+++
141
++c
a+c
+b+
ab+
26
+bc
abc
+++
a++
25
ab+
ab+
++c
++c
2
a+c
abc
+++
+b+
3
a. In how many cells has meiosis occurred to yield these data?
334
b. Give the best genetic map to explain these results. Indicate all relevant genetic distances,
both between genes and between each gene and its respective centromere.
First designate the type of asci present. Remember that this will be done for each pair of genes
(PD (P), NPD (n), and T . When PD = NPD then the genes are not linked.
A – b PD = 137 and NPD = 141 + 2 = 143 PD = NPD therefore not linked.
B – c PD = 137 + 26 + 2 = 165 and NPD = 141 + 25 = 166 PD = NPD therefore not linked.
A – c PD = 137 + 141 = 278 and NPD = 3 PD > NPD therefore linked.
Mu = [(PD + 1/2T)/total ] x 100 [2 + 1.2(26 +25)/334] x 100 = 8.2 cM
= second division segregation * ½ / total
For a + and a + = [(26 + 25)x .5]334 = 0.076 = RF m.u = 7.6
For b + and b + = (3 x 0.5)/334 = 0.004 Rf = 0.4 m.u.
For + c and + c = 0 on top of centromere
c. Diagram a meiosis that could give rise to one of the three tetrads in the class at the far
right in the list.
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