GTS 161 - PRACTICAL 3 – 2014
ADDITIONAL QUESTIONS
A1.
You cross the following individuals with each other, and interbreed the F1:
AAbbCC
x
aaBBCC
a.
What phenotypic ratio do you expect in the F2, if both loci show complete dominance and
sort independently?
b.
You obtain the following F2 progeny:
380
A_B_C_
16
aabbC_
132
aaB_C_
112
A_bbC_
Use a chi-square test to compare these results with those expected for the cross.
TEXTBOOK QUESTIONS CHAPTER 3
Application Questions and Problems: 39 a, b (25 a, b); 41 (27)
39.
In the 1800s, a man with dwarfism who lived in Utah produced a large number of
descendants: 22 children, 49 grandchildren, and 250 great-grandchildren (see the
illustration of a family pedigree), many of whom were also dwarfs (F. F. Stephens. 1943.
Journal of Heredity 34:229–235). The type of dwarfism found in this family is called
Schmid-type metaphyseal chondrodysplasia, although it was originally thought to be
achondroplastic dwarfism. Among the families of this kindred, dwarfism appeared only in
members who had one parent with dwarfism. When one parent was a dwarf, the following
numbers of children were produced.
Family in which
one parent
had dwarfism
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
Total
Children with normal
stature
Children with
dwarfism
15
4
1
6
2
8
4
2
0
3
2
2
2
1
0
52
7
6
6
2
2
4
4
1
1
1
3
1
0
0
2
40
a.
With the assumption that Schmid-type metaphyseal chondrodysplasia is rare, is this type of
dwarfism inherited as a dominant or recessive trait? Explain your reasoning?
b.
Based on your answer for part a., what is the expected ratio of dwarf and normal children in
the families given in the table. Use a chi-square test to determine if the total number of
children for these families (52 normal, 40 dwarfs) is significantly different from the number
expected.
Hint 1: The total of 52:40 offspring numbers indicates a potentially 1:1 ratio.
Hint 2: What kind of cross will produce this?
Hint 3: Remember that if a disease is ‘rare’ in a population, this means the chances are small for
any individual to carry the disease allele, and even smaller that he/she will have 2 copies
of the allele (ie inherited from each parent). Alternatively you would not expect a
significant number of unrelated individuals in the population to have this allele.
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41.
In the California poppy, an allele for yellow flowers (C) is dominant over an allele for white
flowers (c). At an independently assorting locus, an allele for entire petals (F) is dominant
over an allele for fringed petals (f). A plant that is homozygous for yellow and entire petals
is crossed with a plant that is white and fringed. A resulting F1 plant is then crossed with a
plant that is white and fringed, and the following progeny are produced: 54 yellow and
entire; 58 yellow and fringed, 53 white and entire, and 10 white and fringed.
a.
b.
c.
Use a chi-square test to compare the observed numbers with those expected for the cross.
What conclusion can you make from the results of the chi-square test?
Suggest an explanation for the results.
TEXTBOOK QUESTIONS CHAPTER 4
Comprehension Questions: 4; 5; 8 (Not present in 4th Ed)
4.
How does sex determination in the XX-XY system differ from sex determination in the ZZZW system?
5.
What is the pseudoautosomal region. How does the inheritance of traits encoded by genes
in this region differ from the inheritance of other Y-linked characteristics?
8.
Give the typical sex-chromosome complement found in the cells of people with Turner
syndrome, with Klinefelter syndrome, and with androgen-insensitivity syndrome. What is
the sex-chromosome complement of triplo-X females?
Application Questions and Problems: 23 (9); 27 (13); 35 (20); 36 a (21 a); 42 (27)
23.
a.
b.
c.
d.
27.
a.
b.
35.
Joe has classic hemophilia, an X-linked recessive disease. Could Joe have inherited the
gene for this disease from the following persons?
Yes
No
His mother’s mother
___
___
His mother’s father
___
___
His father’s mother
___
___
His father’s father
___
___
Red-green color blindness in humans is due to an X-linked recessive gene. A woman whose
father is color blind possesses one eye with normal color vision and one eye with color
blindness.
Propose an explanation for this woman’s vision pattern. Assume that no new mutations
have spontaneously arisen.
Would it be possible for a man to have one eye with normal color vision and one eye with
color blindness?
Red-green color blindness is an X-linked recessive trait in humans. Polydactyly (extra
fingers and toes) is an autosomal dominant trait. Martha has normal fingers and toes and
normal color vision. Her mother is normal in all respects, but her father is color blind and
polydactylous. Bill is color blind and polydactylous. His mother has normal color vision and
normal fingers and toes. If Bill and Martha marry, what types and proportions of children
can they produce?
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36.
A Drosophila mutation called singed (s) causes the bristles to be bent and misshapen. A
mutation called purple (p) is another recessive mutation that causes the fly’s eyes to be
purple in color instead of the normal red. In the P generation, flies homozygous for singed
and purple were crossed with flies that were homozygous for normal bristles and red eyes.
The F1 were intercrossed to produce the F2 , and the following results were obtained.
Cross 1
P: male, singed bristles, purple eyes X female, normal bristles, red eyes
F1: 420 female, normal bristles, red eyes
426 male, normal bristles, red eyes
F2: 337 female, normal bristles, red eyes
113 female, normal bristles, purple eyes
168 male, normal bristles, red eyes
170 male, singed bristles, red eyes
56 male, normal bristles, purple eyes
58 male, singed bristles, purple eyes
Cross 2
P: female, singed bristles, purple eyes X male, normal bristles, red eyes
F1: 504 female, normal bristles, red eyes
498 male, singed bristles, red eyes
F2: 227 female, normal bristles, red eyes
223 female, singed bristles, red eyes
225 male, normal bristles, red eyes
225 male, singed bristles, red eyes
78 female, normal bristles, purple eyes
76 female, singed bristles, purple eyes
74 male, normal bristles, purple eyes
72 male, singed bristles, purple eyes
a.
b.
42.
a.
b.
What are the modes of inheritance of singed and purple? Explain your reasoning.
Give genotypes for the parents and offspring in the P, F1 , and F2 generations of Cross 1 and
Cross 2.
A woman with normal chromosomes mates with a man who also has normal chromosomes.
Suppose that during oogenesis, the woman’s sex chromosomes undergo nondisjunction in
meiosis I; the man’s chromosomes separate normally. Give possible combinations of sex
chromosomes that might occur in this couple’s children and the number of Barr bodies you
would expect to see in each of the cells of each child.
What chromosome combinations and numbers of Barr bodies would you expect to see if the
chromosomes separate normally during oogenesis, but nondisjunction of the sex
chromosomes takes place in meiosis I of spermatogenesis?
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ADDITIONAL QUESTIONS
A2.
a.
b.
c.
d.
e.
f.
g.
A3.
a.
b.
A4.
a.
b.
c.
What will the X : A ratio and resulting sexual phenotype be of Drosophila flies with the
following chromosome compositions?
(A = one haploid set of autosomal chromosomes; X and Y = sex chromosomes.)
AAX
AAXXY
AAAXX
AAXX
AAXXX
AAAXXX
AAY
Dosage compensation leads to a variety of interesting coat colour patterns in certain
mammals. For instance, a female cat that is heterozygous for the two coat colour alleles
black and orange will have the tortoiseshell or mosaic phenotype.
Describe the chromosomal basis for the mosaicism (tortoiseshell) in the female.
Explain why chromosomally normal male cats do not show the mosaic phenotype, but XXY
male cats can be tortoiseshell.
In chickens a dominant autosomal allele causes the pea comb. Barred feathers are as a
result of a dominant sex linked gene. (Remember in chickens, ♀ is ZW and ♂ is ZZ.)
A rooster and a hen, which both have pea combs and barred feathers, produce a chicken
that has a single comb and non-barred feathers. What are the parental genotypes?
What is the sex of the chicken?
Give the expected genotypic and phenotypic ratios of the offspring from this cross.
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