pl (ch1) 1/30
pl (ch2) 0
pl (ch3) 2/30
pl (CH3) 2/30
Number of alleles = 4
Alleles frequency = ¼
PI = 2 x ((1/4)81/4)81/30)) + (1/481/4) 8 2/30) + (1/481/4)82/30
b) It has different results due to the difference in population
2.
| DD 38 | 76 D alleles |
| EE 20 | 40 E alleles |
| FF 4 | 8 F alleles |
| DE 16 | 16 D & 16 E alleles |
| DF 14 | 14 D & 14 F alleles |
| EF 8 | 8F alleles & 8 E alleles |
| Total alleles | 200 |
| Total progeny | 100 |
Frequency of D alleles = 106/200 = 0.53
Total D alleles = 14 +16+ 76= 106
Total E alleles = 8+16+ 40 = 64
Frequency E alleles= 64/200 = 0.32
Frequency F alleles= 30/200 = 0.15 given total E alleles= 14+8+8 = 30
b)
Expected genotype number:
DD = 0.53 x 0.53 x 100 = 28.09
EE = 0.32 x 0.32 x 100= 10.24
DE= 0.53 x 0.32 x 100 x 2 = 33.92
DF = 0.15 x 0.53 x 2 x 100 = 15.9
EF = 0.15 x 0.32 x 2 x 100= 9.6
c) yes
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