1. If a population comprised 52 AA, 114 Aa and 34 aa individuals and the ‘A’ all
ID: 273952 • Letter: 1
Question
1. If a population comprised 52 AA, 114 Aa and 34 aa individuals and the ‘A’ allele was dominant, what would be the
allele, genotype and phenotype frequencies (expressed as a percentage)?
Hint: Since we have given you the genotypes, you can see that this is not a Hardy-Weinberg situation
2. In Practical 8, you put 5 male and 5 female wild type (vg+/vg+) flies, with 5 male & 5 female vestigial winged (vg-/vg-)
flies into your control bottle.
Assuming equal fitness, what would you expect the phenotype ratio to be (wild type to vestigial)?
Hint: Look at all the possible mating events (wild type male to wild type female, wild type male to vestigial winged
female, etc.) that could take place and draw Punnet squares for each
Explanation / Answer
1. The answer is
1. Allele frequency estimation:
Genotype
Freequency
Allele A
Allele a
Total
AA
52
104
0
104
Aa
114
114
114
228
aa
34
0
68
68
Total
200
218
182
400
Allele frequency = Frequency of a allele/Frequency of all allales
Allele
Count
Allele frequency
Percentage
Allele A
= 218/400
0.55
55% (0.55*100)
Allele a
= 182/400
0.45
45%
(0.45*100)
2. Genotype frequency estimation
Genotype
Frequency
Percentage
AA
= 0.55*0.55 = 0.30
30% (0.30*100)
Aa
= 2*0.45*0.55 = 0.50
50% (0.50*100)
aa
= 0.45*0.45= 0.20
20% (0.20*100)
3. Phenotype frequencies:
Dominant allele can express her character in both homo and heterozygous condtions, so, the A phenotype counts both AA and Aa.
The percentage of dominant phenotype = A_(AA+Aa)= 30+50 = 80%
The percentage of recessive phenotype = aa = 20%
ACCORDING TO CHEGG GUIDELINES WE HAVE TO ANSWER ONE QUESTION AT A TIME. POST THE REST AS SPERATE QUESTIONS, THEN I CAN HELP YOU.
Genotype
Freequency
Allele A
Allele a
Total
AA
52
104
0
104
Aa
114
114
114
228
aa
34
0
68
68
Total
200
218
182
400
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