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Problem Set 8, Spring 2017
Name:
8 points total
Student ID:
1. The mythical Antarctic toad is found in several true breeding strains. One has blue skin and long toes.
The second has red eyes. You cross blue-skinned, long toed X red eyed and find the progeny all have all
wild type phenotypes (i.e. not blue skinned, long toed or red eyed). You test cross these F1’s and find the
results shown below. Only the recessive phenotypes are listed below
123 long toed, blue skinned, red eyed
4513 long toed, blue skinned
a. What do the results shown above say about
118 blue skinned, red eyed
the chromosomal location of the three genes?
19 long toed, red eyed
125 long toed
They are linked
4781 red eyed
21 blue skinned
134 wild type
9834
b. Please calculate the genetic distance between the l and b genes, i.e. the
genes that produce long toes and blue skin.
l – b: (118 + 125 + 19 + 21)/9834 X 100% = 2.9 cM
c. There are three linked genes, a, b and c. a and b are 10 map units apart and b and c are 20 map units
apart and a and c are 30 map units apart. If you cross a b c/+ + + X the triple recessive, what fraction of
the progeny will be phenotypically a- (i.e. homozygous a-) and b- but wild type for c? (hint: the
probability of doubles = the product of the probability of singles)
This is a simple three factor mapping problem, but worked backwards. 30% of the offspring will be
recombinant because the outer genes, a and c are 30 map units apart. Of these, 0.1(0.2) = 0.02 will
be doubles. To produce the desired genotype, a b +/a b c, we need a single cross over between b and
c, but not double cross over (because that will produce a + c and + b + chromosomes). So the answer
is ½(0.2 – 0.02) = 0.09 or 9%.
2. Several strains of the seabass B. kochae, have accumulated extensive rearrangements of chromosome 2,
as shown below.
original chromosome
strain I
strain II
strain III
Determine the order in which the changes had occurred and name the type of rearrangement.
(note: each new strain resulted from a single rearrangement in its parent.)
original
pericentric inversion
strain 2
deletion
strain 1
paracentric inversion
strain 3
1
Problem Set 8, Spring 2017
8 points total
Name:
Student ID:
3. Name the chromosomal anomoly present for each pair of chromosomes (the wild type chromosome is
the uppermost chromosome of each pair, also indicated by the star). Also, show the chromosomes as they
would appear synapsed in meiosis I, including letters and centromeres.
a.
u
v
w
x
y
z
u
v
w
x
y
z
w
v
v
This is a duplication
(actually a triplication
in this case).
w
b.
This is a reciprocal
translocation.
z
z
u
v
w
x y
y d
c
b
a
u
v
w
x
e d
f
c
b
a
e
f
4. You map two genes, bl, which causes blue fur and grn, which causes green teeth, and find that they are
21 mu apart. Starting with another pair of parents (one heterozygous and the second homozygous
recessive) you find the following results:
817
793
37
42
blue fur, green teeth
wild type
blue fur, normal teeth
normal fur, green teeth
How can you explain this outcome?
37 + 42
1689
= 4.7 cM
Maybe there’s an inversion or deletion
between the two genes that encompasses
~16.3 cM.
2
Problem Set 8, Spring 2017
Name:
8 points total
Student ID:
5. In studying a particular animal, you find that some of the chromosomes appear a bit unusual. . The
relevant chromsomes are shown below. One pair of homologues is on the left and the second pair is
shown on the right. For each pair, the normal chromosome is the one on top.
a. What is the name of the chromosomal alteration that generated the abnormal chromosomes?
Reciprocal translocation
b. Show the chromosomes synapsed in meiosis. Please include the genes.
c. You find an individual homozygous for the abnormal chromosomes as shown below. Show this
individual's chromosomes synapsed in meiosis. Please include the genes.
The answer is to draw the chromosomes essentially as they are given.
3
Problem Set 8, Spring 2017
Name:
8 points total
Student ID:
6a. A woman with Turner syndrome (XO) is found to be color-blind. Both her mother and father have
normal vision. How can her colorblindness be explained? Did the nondisjunction occur in the father or
mother?
Mom had to carry Xcb. Nondisjunction in dad produced gamete lacking X and Y, which
fused to Xcb gamete from mom.
b. A man with Klinefelter syndrome (XXY) is found to be color-blind. Both his mother and father have
normal vision. How can his colorblindness be explained? Did the nondisjunction occur in the father or
mother?
Nondisjunction in meiosis II of mom produced gamete that was XcbXcb, which fused with Ycontaining gamete from dad.
7. While looking at a large number of siblings that are all homozygous wild type for the prpl gene (which
causes purple blotches when homozygous mutant), you find that each individual has a different pattern of
purple blotches on their skin. How might you explain this result, assuming that it results from an effect on
the prpl gene?
This might result from position effect variegation. Perhaps a chromosomal rearrangement in
that family has put the prpl gene near a heterochromatic region.
8. Strains heterozygous for each of six Drosophila mutations, a – f, and for each of six different deletions,
1 – 6 are generated. The table below shows whether such strains displayed wild type (+) or mutant (-)
phenotypes. The figure illustrates the region deleted in each deletion strain (region deleted indicated by a
bar) as determined by cytological examination of polytene chromosomes.
a
b
c
d
e
f
del 1 +
–
+
+
+
–
del 2 +
–
+
+
–
–
del 3 –
+
–
+
+
–
del 4 –
+
–
–
+
+
del 5 –
+
–
+
+
+
del 6 –
–
–
+
+
–
Assign each recessive mutation to the relvant region of the chromosome.
e
b
f
a, c
4
d