1. Female: AO, since father had a B phenotype, he must have contributed an i allele.
Male: AO, since mother must have donated an i allele.
Possible genotypes, therefore would be A or O.
Ans: two of the above are correct (e)
2. Dominance Relationships: C>ck>cd>ca
Sepia x cream: since the sepia parent had an albino parent (caca) its genotype must be
ckca. The cream individual had two sepia parents- one of those parents could be either
ckcd OR ckca.
One cross would therefore be:
ckca x cdcd
½ cd
½ ck
½ ckcd (sepia)
½
cd
½ cd
½ ca
½ cacd (cream)
½
cd
Ans: 1 sepia: 1 cream (a)
The other cross would therefore be:
ckca x cdca
½ cd
½ ck
2/4 ckcd (sepia)
½ ca
½ cd
1/4 cacd (cream)
½ ca
½ ca
1/4 caca (albino)
Ans: 2 sepia: 1 cream: 1albino
I will count both (a) and (e) on personal quizzes.
3. Sepia x cream: since the sepia parent had two full color parents (Cck x Cck,d,a) its
genotype could be ckck,d, or a. Again, the cream individual had two sepia parents- one of
those parents could be either ckcd OR ckca.
The results could be the same as 2, and in addition, all sepia (if first parent was ckck ) I
will count both (a) and (e) on personal quizzes.
4. These results are a variation on what would be expected with a monohybrid F1 cross,
or a testcross. Since palomino does not breed true, it does not seem to be due to a specific
allele, and cross #3 argues for incomplete dominance:
Chestnut x cDcD = chestnut
Chestnut x cdcd = cremelo
Chestnut x cDcd = palomino
cDcd x cDcd = ¼ cDcD: ½ cDcd: ¼ cdcd
Ans: incomplete dominance (b)
5. Cross number 3 suggests that this cross is an example of 4 distinct phenotypes arising
from a dihybrid cross, since it is a variant 9:3:3:1 ratio. The gray individuals would be
homozygous at one locus (AAbb) the brown individuals would be homozygous at a
second locus (aaBB) and individuals which inherit at least one dominant allele at both
loci would be green (A_B_). Individuals that are homozygous recessive at both alleles
(aabb) would be blue. Since the F2 ratio is a variant 9:3:3:1 ratio, the F1 must be
heterozygotes (AaBb).
Since one of the cats was gray (e.g. AAbb), the other parent must have been brown
(aaBB).
Ans: AAbb x aaBB; (c)
6. The parental classes would be the most
abundant:
(pr + +) and (+ v bm)
The double recombinant classes would be the
least abundant:
(+ + +) and (pr v bm)
Therefore, the middle allele must be purple:
+ pr
+
v +
bm
Ans: pr (a)
7.
v-pr interval:
v pr +
+ + bm
+++
v pr bm
79
82
42
44
247
247/1109 = 0.223
22.3 map units
Phenotype
Number of Progeny
virescent
200
purple
237
brown
82
virescent, brown
230
virescent, purple
79
purple, brown
195
wild type
42
virescent, purple, brown 44
pr-bm interval:
+ pr bm
195
v++
200
+++
42
v pr bm
44
481
481/1109 = 0.434
43.4 map units
Ans: 43.4 map units (c)
v- bm interval: 22.3 + 43.4 = 65.7 map units.
8. Observed DCOs: 42+44 = 86
Expected DCOs: (0.434*0.223*1109) = 107.3
86/107 ≈ 0.803 = coefficient of coincidence
1-0.803 = 0.196
Ans: (b)
Spore
number
1
2
3
4
5
6
7
8
totals
type
segregation
-------------------------------------------------Ascus Types------------------------------------------
1
2
3
4
5
6
7
++
++
ab
ab
+b
+b
a+
a+
5
TT
2:2
++
++
ab
ab
++
++
ab
ab
1
NPD
2:2
+b
+b
ab
ab
++
++
a+
a+
5
TT
2:1
a+
a+
a+
a+
+b
+b
+b
+b
808
PD
1:1
9. 5+1+5+90= 101/1000* ½ = 0.0505 = 5.05 map units
Ans: (a)
10. RF between a and b = (½ TT+ NPD)/total
½ (5 + 5 + 90) + 2 = 52
52/1000 = 0.052 = 5.2 map units
Ans: (b)
+b
+b
a+
a+
+b
+b
a+
a+
90
PD
2:2
++
++
++
++
ab
ab
ab
ab
1
NPD
1:1
++
++
+b
+b
a+
a+
ab
ab
90
TT
1:2