Homework Answers
Answer:
a).
Tt x Tt ---Parents
|
T |
t |
|
|
T |
TT (Tall) |
Tt (tall) |
|
t |
Tt (tall) |
tt (short) |
Frequency of tall = ¾ = 75%
Frequency of short = ¼ = 0.25
b).
Expected tall = ¾ * 50 = 37.5
Expected short = ¼ * 500 = 12.5
c).
|
Phenotype |
Observed(O) |
Expected (E) |
O-E |
(O-E)2 |
(O-E)2/E |
|
tall |
30 |
37.5 |
-7.5 |
56.25 |
1.50 |
|
short |
20 |
12.5 |
7.5 |
56.25 |
4.50 |
|
Totals |
50 |
50 |
6.00 |
Chi-square value = 6.00
Degrees of freedom= number of phenotypes – 1
Df = 2-1 = 1
The p value = 3.84.
The chi-square value of 6.00 is greater than the p-value of 3.84. Hence the hypothesis is rejected and the data is not fit in the hypothesis.
d). Degrees of freedom= number of phenotypes – 1
Df = 2-1 = 1
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