Genetics 314 - Spring 2009
Homework #3 - 25 points
Due Wednesday, April 22th
1.
Looking at a cross involving a true breeding awned club wheat and a true-breeding awnless common
wheat you get the following data:
F1 - all awnless and club head type
F2 - four phenotypes:
awnless, club
awned, club
awnless common
awned, common
179 plants
64 plants
67 plants
15 plants
a) Which traits are dominant and which traits are recessive?
Based on the F1 phenotypes, awnless and club head type are dominant
b) What ratio would you expect if this was a two gene model with both genes showing complete
dominance?
9 : 3 :3 : 1 ratio
c) Test your hypothesis using Chi square analysis.
ob
179
exp
325/16 x 9 = 183
ob-ex
179-183 = -4
(ob-exp)2
16
(ob-ex)2/exp
16/183 = .087
64
325/16 x 3 = 61
64-61 = 3
9
9/61 = .15
67
325/16 x 3 = 61
67-61 = 6
36
36/61 = .59
15
325/16 x 1 = 20
20-15 = 5
25
25/20 = 1.25
χ2 = 2.077
df = (n – 1) = (4 – 1) = 3
at 3 df at 0.05 Chi square value is 7.84
2.077 is less than 7.84 so accept hypothesis
2.
You make a cross between the following two genotypes:
Aa Bb Cc Dd Ee Ff x Aa BB CC Dd ee Ff
Out of 4096 progeny how many progeny would have the following genotypes:
A
B
C
D
E
F
a) AA BB CC DD ee ff .25 x .5 x .5 x .25 x .5 x .25 x 4096 = 8
b) Aa Bb Cc Dd Ee Ff . 5 x .5 x .5 x .5 x .5 x .5 x 4096 = 64
c) Aa bb Cc dd ee Ff
.5 x 0
x 4096 = 0
d) AA Bb CC Dd Ee ff 25 x .5 x .5 x .5 x .5 x .25 x 4096 = 16
e) Aa BB Cc DD Ee ff .5 x .5 x .5 x .25 x .5 x .25 x 4096 = 16
3.
You are asked by a friend to calculate the probabilities of having the following combination of
children:
a)
b)
c)
d)
a)
b)
c)
d)
4.
7 boys - 2 girl
5 boys - 4 girls
3 boys - 6 girls
0 boys - 9 girls
9!/7!2!x.57x.52 = 0.07
9!/5!4!x.55x.54=0.25
9!/3!6!x.53x.56=0.164
9!/0!9!x.50x.59=0.00195
What is the probability of having 5 boys - 4 girls in a specific order (example: boy, girl, girl, boy, girl,
girl, boy)? Does it differ from your answer to 3b? Why or why not?
The probability of having 5 boys and 4 girls in a specific order is .5 9 = 0.00195. This does differ from
your answer in 3b because there you were looking at all the possible combinations of 5 boys and 4
girls whereas here you are only looking at one possibility.
5.
You are studying eye color in Fruit flies. You discover flies with slightly different shades of white eyes
where the normal (wild type) has red eyes. You wonder if you are observing allelic differences or if the
difference is due to different genes.
a) What type of cross would you make to determine the genetic control of the eye color
b)
c)
difference?
What results would you expect if the difference was due to different alleles? Why?
What results would you expect if the difference was due to different genes? Why?
a) You would make a complementation cross.
b) You would expect the progeny to have white eyes
white
white
a1a1BB
X
a2a2BB
1 2
a a BB
white
c) You would expect the progeny to have red eyes
white
white
aaBB
X
AAbb
AaBb
red
6.
You are studying skin color in snakes. You mate a true breeding brown snake with a yellow snake and
all the F1 progeny are black. You mate the F1 snakes and recover the following progeny:
phenotype
black
brown
yellow
total
no. of progeny
41
12
19
72
a) What type of gene action are you observing?
b) Test your hypothesis using Chi square analysis.
a) you are observing single recessive epistasis (9 : 3 :4)
b)
ob
41
12
19
exp
(72/16)*9=40.5
(72/16)*3=13.5
(72/16)*4=18
(ob-exp)2
(41-40.5)2=0.25
(12-13.5)2=2.25
(18-19)2=1
(ob-exp)2/exp
0.25/40.5=0.006
2.25/13.5=0.167
1/18=0.056
χ2
0.229
df=(n-1)=(3-1)=2
with 2 df at 0.05 level χ2 is 5.99
0.229 is less than 5.99 so accept the hypothesis that this is a 9:3:4 ratio of single recessive epistasis
7.
You are studying mouse coat color. You mate a true breeding brown mouse with an albino mouse and
all the F1 progeny are black. You mate the F1 mice and recover the following progeny:
phenotype
no. of progeny
black
brown
albino
total
160
60
80
300
a) What type of gene action are you observing?
b) Test your hypothesis using Chi square analysis.
a)
You are observing single recessive epistasis (9 : 3 : 4)
b)
ob
160
60
80
exp
(300/16)*9=169
(300/16)*3=56
(300/16)*4=75
(ob-exp)2
(160-169)2=81
(60-56)2=16
(80-75)2=25
(ob-exp)2/exp
81/169=0.479
16/56=0.286
25/75=0.333
χ2
1.095
df=(n-1)=(3-1)=2
with 2 df at 0.05 level χ2 is 5.99
1.095 is less than 5.99 so accept the hypothesis that this is a 9:3:4 ratio of single recessive epistasis
8.
Three traits (colored, full, starchy) have been found to be all on chromosome 9 in corn. A cross was
made between two true breeding parents that differed for the three traits and then the F 1 was crossed to
a completely homozygous recessive (colorless, shrunken waxy) individual. You get the following
results:
colored, full, starchy
colorless, shrunken, waxy
colored, shrunken, waxy
colorless, full, starchy
colorless, full, waxy
colored, shrunken, starchy
colored, full, waxy
colorless, shrunken, starchy
total
17,625
17,375
510
490
3,050
3,250
32
28
42,360
a) What were the phenotypes of the two true breeding parents? What do you base your
conclusion on?
b) What is the gene order?
c) What are the map distances?
d) What is the interference value?
a) Parents are the same as the phenotypes with the largest numbers: they are colored, full,
starchy and colorless, shrunken, waxy.
b) Gene order is: colored----starchy----full
c) C---2.5 mu---S---------15 mu-----------F
C-S
S-F
(490+510+32+28)/42360=0.025*100=2.50
(3050+3250+32+28)/42360=0.15*100=15
d) 0.025*0.15*42360=159 expected double cross-over events
32+28=60 observed double cross-over events
1-(observed dco/expected dco)= interference value
1-(60/159)=0.623