1. Males ZZ; females, ZW
Nonbarred males x barred females
bb x BW
b
b
B
W
Bb
bW
Bb
bW
Ans: a) daughters all one type, sons all the other type
2. Only males are affected: first assume X linkage:
XsY x X+X+ → XsX+ females and X+Y (all short tailed.
X+
Xs
X+
Y
XsX+ XsY
X+X+ X+Y
Note: you would expect ½ of the males to be longed tailed NOT ¼!
Only males are affected: suppose the gene was autosomal recessive
ss males x SS females → Ss females and Ss males (all short tailed).
S
s
S
s
SS
Ss
Ss
ss
Here: you would expect ¼ of the progeny to be longed-tailed...if the trait were not
expressed in females (sex-limited) then all of the females would be shorttailed, but ¼ of
the males long tailed.
Ans: (c) autosomal recessive with sex-limited expression.
3. There is no variability in the expression of the traits discussed...tribbles are either
green or pink.
We are told that G is dominant at a single locus. G_ individuals should be green. If two
true breeding lines are crossed, the genotype of the F1 should be Gg, but only 90% of this
phenotype are green.
Ans: (d) 90% penetrance of the G allele in the heterozygote
4.
Observed
Expected
(O-E)
(O-E)2
(O-E)2/E
54
50
4
16
0.32
55
50
5
25
0.5
45
50
-5
25
0.5
46
50
-4
16
0.32
200
1.64 =c2
Degree of freedom = 3
Let independent assortment be the null hypothesis
If so, then we would expect equal numbers in each category
The chi square value is 1.64, at three degrees of freedom.
This value is CONSISTENT with the hypothesis that the genes are segregating
independently- they are far enough apart on the X chromosome that the appear to be
unlinked.
Ans: (b) 0.9>p>0.5; no linkage
5. Females:
All wild type:
1000 progeny
Scute sable vermilion
Sable vermilion
Scute vermilion
Vermilion
Sable
Scute sable
Scute
Wild type
290
149
30
10
46
14
157
304
Males:
From examining the most common classes (the nonrecombinant chromosomes from the
heterozygote) and the least common classes (the double recombinant chromosomes) we
conclude that all wild type alleles are present on one homolog, all mutant alleles are
present on the other homolog, and the vermilion locus is in the middle, because it has
changed orientation in the dco relative to the phasing arrangement found in the
nonrecombinant chromosomes:
+
sc
sc-v interval:
sc
157
v sa 149
v
10
sc sa 14
330/1000 = 0.33 = 33 map units
+
v
+
sa
v-sa interval:
sa
46
sc-v 30
v
10
sc sa 14
100/1000 = 0.10 = 10 map units
sc-v interval: Ans: (d) 33 m.u.
6. To solve this problem, we essentially need to work backwards. There will be eight
phenotypic classes generated from the cross as depicted, a pair of non-recombinant or
parental chromosomes, two pairs of single cross-over classes, and a pair of double cross
over classes:
sc -------------33-------- v ----10---  sa
sc
+
+
v
+
X
sa
sc + +
+ v sa
nonrecombinant (parental)
nonrecombinant (parental)
sc v sa
+ ++
single crossover between sc and v
single crossover between sc and v
sc + sa
+v+
single crossover between v and sa
single crossover between v and sa
sc v +
+ + sa
double crossover
double crossover
It is the frequency of the + v + class that we are asked to calculate.
From the previous problem, we expect 10% of the meiotic products to show a crossover
in the vermilion - sable gene interval. The meiotic products representing a crossover in
this interval will include both single and double crossover classes. Moreover, from the
previous problem, there is evidence of interference, because (0.33)(0.1)(1000) = an
expected value of 33 DCOs, but only 24 were detected. This is a coefficient of
coincidence of 0.74- only 74% of the expected double crossovers were observed.
Let’s first calculate the results expected for the DCO classes. If we were screening 1500
progeny, the number we would expect in this category would be (0.33)(0.1)(0.74)(1500)
= 36.6 progeny. Therefore, between 18-19 progeny would be expected in each DCO
class.
Again, the frequency of crossovers in the v sa interval represents the total number of
crossovers, so the number of SCOs would be [1500 (0.1)] – DCO class. This would
represent 113 individuals in both classes, or 57 individuals in the + v + class. Note that
57 (sc + sa) + 57 (+ v +) +18 (sc v +) +18(+ + sa) = 150/1500 = 0.1
Ans: (e) none of the above. (I counted (a) as correct on the quiz, but you should be able
to go through this logic. Also see problem 11, chapter 12, in Thompson.
7. Affected parent gives rise to affected children; about ½ of children are affected,
regardless of sex.
Ans: True, these are characteristics of an autosomal dominant trait.
8.
If we look at the progeny, the nail patella disease gene segregates with the IB locus, and
the i gene segregates with the wild type locus in 15/18 progeny. This is greater than the
1:1:1:1 distribution we would predict on the basis of independent assortment.
OD
obs
exp
(O-E)
(O-E)2
(O-E)2/ E
BD
1.00
4.50
-3.50
12.25
2.72
O+
8.00
4.50
3.50
12.25
2.72
B+
7.00
4.50
2.50
6.25
1.39
2.00
4.50
-2.50
6.25
1.39
18.00
18.00
8.22 >7.81
RF = 3/18 = 0.167 = 16.7 m.u.
Now suppose individual III-1 were crossed to a normal woman with type O blood: What
would be the chances of producing a child with nail patella syndrome and type O blood?
IB
NP
X
X
i
+
i
NP
i
+
i
i
+
+
8. 0.167/2 = 0.083 Ans: (b)
Spore
number
1
2
3
4
5
6
7
8
totals
-------------------------------------------------Ascus Types-------------------------------------------
1
2
3
4
5
6
7
ab
ab
+b
+b
a+
a+
++
++
90
ab
ab
++
++
+b
+b
a+
a+
10
+b
+b
+b
+b
a+
a+
a+
a+
1
++
++
++
++
ab
ab
ab
ab
699
a+
a+
a+
a+
+b
+b
+b
+b
5
ab
ab
+b
+b
++
++
a+
a+
190
ab
ab
++
++
++
++
ab
ab
5
Identify the number of second division segregation patterns for the B gene:
10 + 5 = 15
(15/1000) x ½ = .75 map unit
9. Ans: (e) none of the above
10. From the above, the tetratype (TT) classes are (1 = 90), (2 = 10) and 6 (190)
The nonparental ditype (NPD) classes are (3 = 1), (5 = 5), so TT = 290 and NPD
= 6.
RF = (½ TT + NPD)/total asci ≈ 0.151 = 15.1 map units (book formula)
RF = (½ TT + 3 NPD)/total asci ≈ 16.3 map units (class formula)
Ans: (c); 16 map units.