Download GENETICS Homework 3

Fall 2005
Due: 10/5
GENETICS
Homework 3
1. (1.5 pts.) Given the following gene-to gene distances, determine a map that includes
all genes.
O-R
M-R
M-G
M
4
3
7
12
O
R-A
G-A
G-N
3
13
8
10
R
R-G
O-G
O-N
5
5
8
18
G
8
A 2 N
2. (1.5 pts.)The following two genotypes are crossed: Aa;Bb;Cc;dd;Ee X
Aa;bb;Cc;Dd;Ee. What will the proportion of the following genotypes be among the
progeny of this cross?
a)
Aa:Bb;Cc;Dd;Ee
Pr = (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32
b)
Aa;bb;Cc;dd;ee
Pr = (1/2)(1/2)(1/2)(1/2)(1/4) = 1/64
c)
aa;bb;cc;dd;ee
Pr = (1/4)(1/2)(1/4)((1/2)(1/4) = 1/256
d)
AA;BB;CC;DD;EE
Pr = 0
3.
(1.5 pts.) In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and
yellow spots (S) are dominant over no spots (s). The genes for these two
characteristics assort independently. A homozygous plant that has bitter fruit and
yellow spots is crossed with a homozygous plant that has sweet fruit and no spots.
The F1 are intercrossed to produce the F2.
a) What will be the phenotypic ratios in the F2?
9/16 B_S_
3/16 B_ss
3/16 bbS_
1/16 bbss
bitter with yellow spots
bitter with no spots
sweet with yellow spots
sweet with no spots
b) If an F1 plant is backcrossed with the bitter, yellow spotted parent, what
phenotypes and proportions are expected in the offspring?
All would be bitter with yellow spots
c) If an F1 plant is backcrossed with the sweet, nonspotted parent, what
phenotypes and proportions are expected in the offspring?
¼ bitter with yellow spots
¼ bitter with no spots
¼ sweet with yellow spots
¼ sweet with no spots
4. (1.5 pts.) In rabbits, unspotted (W) is dominant to spotted(w) and short (S) is
dominant to long (s). Females heterozygous for both genes are mated to
homozygous recessive males to yield:
75 spotted, long
66 unspotted, short
10 unspotted, long
7 spotted, short
a) Are the two genes linked?
Yes.
b) What is the chromosome composition of the female?
ws/WS
c) If the genes are linked, determine the map distance.
RF = 17/158 = .1076 or 10.76 map units
5. (2 pt.) In assessing data that fell into two phenotypic classes, a geneticist observed
2
values of 250:150. She decided to perform a X analysis by using the following
two different null hypotheses: (a) the data fit a 3:1 ratio, and (b) the data fit a 1:1
2
ratio. Calculate the X values for each hypothesis. What can be concluded about
each hypothesis?
Hypothesis one: expected 300 and 100
X2 = (300-250)2 + (100-150)2 = 8.33 + 25 = 33.33 df = 1 p < 0.005
300
100
REJECT HYPOTHESIS
Hypothesis two: expected 200 and 200
X2 = (200-250)2 + (200-150)2 = 12.5 + 12.5 = 25 df = 1 p <0.005
200
200
REJECT HYPOTHESIS
6. (2 pt.) Consider the three linked genes in corn: b, lg, and v. A test cross between a
triple heterozygote and a homozygous recessive yields the following progeny:
165
37
64
11
9
33
56
125
+
+
b
+
b
b
+
b
v
+
+
+
v
v
v
+
lg
lg
lg
+
lg
+
+
+
Determine the genetic composition of the heterozygote: determine gene order and
map distances: and calculate interference.
+ v lg
b + +
v
18 mu b
28 mu
v-b = 70 + 20 = .18 or 18 m.u.
500
lg
b-lg = 120 + 20 = .28 or 28 m.u.
500
predicted dco = 0.18 x 0.28 x 500 = 25.2
I = 1 – 20/25.2 = 1 - 0.794 = 0.206