a. Is the snapdragon population in Hardy-Weinberg equilibrium? (circle one) yes
ID: 20405 • Letter: A
Question
a. Is the snapdragon population in Hardy-Weinberg equilibrium? (circle one) yes no can’t tell Explain your answer. If calculations are necessary, then you must show them to get full credit. But your explanation must include words, not just calculations.
b. Is the pea population in Hardy-Weinberg equilibrium? (circle one) yes no can’t tell Explain your answer. If calculations are necessary, then you must show them to get full credit. But your explanation must include words, not just calculations.
Explanation / Answer
if there is complete dominance , the ratio will be 3:1
IT IS TESTED BY USING CHI-SQUARE TEST
Observed
Expected
Difference
(O-E)
(O-E)^2/E
p^2
homozygous dominant
91
75
16
3.41
2pq
heterozygous
q^2-
homozygous recessive
9
25
-16
10.24
100
100
13.65
CALCULATED VALUE IS NORE THAN TABLE VALUE AT 5% OF LEVEL OF SIGNIFICANE
HENCE PEAS POPULATION IS NOT IN HARDY WEINBERG EQUILIbrium
=====================
if there is incomplete dominance , the ratio will be 1:2:1
Observed
Expected
Difference
(O-E)
(O-E)^2/E
p^2
homozygous dominant
42
25
17
11.56
2pq
heterozygous
22
50
-28
15.68
q^2-
homozygous recessive
36
25
11
4.84
100
100
32.08
CALCULATED VALUE IS NORE THAN TABLE VALUE AT 5% OF LEVEL OF SIGNIFICANE
HENCE SNAPDRAGON POPULATION IS NOT IN HARDY WEINBERG EQUILIbrium
Observed
Expected
Difference
(O-E)
(O-E)^2/E
p^2
homozygous dominant
91
75
16
3.41
2pq
heterozygous
q^2-
homozygous recessive
9
25
-16
10.24
100
100
13.65
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