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Can u spot the difference in the sentences above?
Actual questions:
Can u spot the difference in the sentences above?
Actual questions:
2 answers:
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Answer:
a) 1/16, b) 4/16, c) 1/16, d) 1/16, e) 4/16, f) 4/16, g) 0 + alleles, h) 1/256, i) 1 + alleles, j) 4/256, k) 2 + alleles, l) 16/256, m) 3 + alleles, n) 24/256, o) 4 + alleles), p) 36/256, q) 5 + alleles, r) 24/256, s) 6 + alleles, t) 16/256, u) 7 + alleles, v) 4/256, w) 8 + alleles, x) 1/256.
You will find the uncompleted Punnett square (missing in the problem statement) and the complete modified Punnett square in the attached files
Explanation:
Due to technical problems, you will find the complete explanation in the attached files
The Hardy-Weinberg equation is as follows:
Where:
(convert all % to decimals)
p= homozygous dominant
q= homozygous recessive
pq= heterozygous
While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.
Using the 1st equation, we can solve for the other dominant/recessive frequency:
1-0.2=0.8
Meaning that:
p= 0.8 & q=0.2
If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.
2(0.8)(0.2)= 2(0.16)= 0.32
So, the heterozygous frequency would be:
0.32
Hope this helps!
Where:
(convert all % to decimals)
p= homozygous dominant
q= homozygous recessive
pq= heterozygous
While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.
Using the 1st equation, we can solve for the other dominant/recessive frequency:
1-0.2=0.8
Meaning that:
p= 0.8 & q=0.2
If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.
2(0.8)(0.2)= 2(0.16)= 0.32
So, the heterozygous frequency would be:
0.32
Hope this helps!