Download click here

1. Independent assortment is based on the fact that the genes are NOT linked. In a
dihybrid cross, you would expect a 9:3:3:1 ratio if genes are not linked. The three ratios
shown are all expected results of a dihybrid (AaBb x AaBb) cross- all show a 9:3:3:1
ratio, or a variant of it.
Ans: all of the above (d)
2. If both A and B are required for flower color, then in an F1 X F1 dihybrid cross, the
only class that would inherit at least one copy of both genes would be:
9/16 A_B_
All other classes (3/16 A_bb; 3/16 aaB_; 1/16 aabb) would be colorless.
Ans: 9:7 (b)
3. In this cross, three genes, not two, are necessary for color (BbDdRr x BbDdRr)
Therefore, the colored individuals must receive a dominant allele from each locus¾ B_ x ¾ D_ x ¾ R_ = 27/64 colored
Everything else is not colored = 37/64
Ans: 27:37 (d)
4. “An inherited state that is determined by segregating alleles of many different genes,
whose interactions (environmental/ genetic)...produce a phenotype exhibiting many
intermediate forms...
This is a strict definition of a continuous or quantitative trait; see pages 68-70 (end of
chapter 3).
Ans: quantitative trait (a)
5. An individual with blood type 0 (ii) will produce antibodies to the antigens found on
the red blood cells of individuals carrying the (IA and IB) alleles.
Although an individual with 0 blood doesn’t make the A and B antigens, other surface
antigens (e.g. M/N, Rh, H) are found on the surface of their cells.
Ans: only two of the above are correct (e) N.B. As per our discussion in class, because
of the ambiguity regarding RBC surface antigens in the book, (d) was also counted as
correct on the quiz- but it’s not! See http://www.bloodbook.com/type-sys.html
6.
XcXc
XhY
XcXh
X
c
+
3 m.u.
+
X
X
h
3% of the meiotic products would be expected to be cross overs... ½ wild type, ½ double
mutant
Ans: 0.03/2 = 0.015 (e)
px sp cn
px sp +
px + cn
px + +
+ sp cn
+ + cn
+++
total
7. What are the most common classes?
px sp +
3497
+ + cn
3482
What are the rarest classes?
px + cn
1
+ sp +
0
sp
px
+
+
cn
X
+
1,461
3,497
1
11
9
3,482
1,539
10,000
px is the middle gene
Ans: True (a)
8.
sp + cn
+ px +
+ px cn
sp + +
Total
9
11
1
0
21
21/10,000 = 0.0021 or 0.21 map units
Ans: (d)
9.
Are the three genes linked?
If so, we would not expect independent assortment; but the K and I loci, and the K and V
loci, appear to assort independently; i.e. a 1:1:1:1 ratio is observed among progeny in
terms of gamete recovery:
Ki
214 + 30
KV
214 + 26
KI
26 + 218
Kv
30 + 218
kI
28 + 222
kV
226 + 28
ki
226 + 36
kv
36 + 222
But the V and I genes do not assort independently:
Vi
214 + 226 = 440
VI
26 + 28 = 54
V
i
vI
218 + 222 = 440
vi
30 + 36 = 66
v
I
54 + 66 = 120/1000 = 0.12
Distance = 12 m.u.
Ans: (c)
X
K
k
k
K
A locus
B locus
PD
1
1
TT
1
2
TT
2
1
PD
2
2
TT NPD NPD
2
2
1
2
2
1
AB x ab
PD>>NPD
A and B are linked
Loci are on opposite sites of centromere
RF = (½ TT + NPD)/total = 32 map units
A to centromere = 22 map units
B to centromere = 14 map units
Ans: greater than 32 m.u. (d)