Genetics! Foot webbing color in geese is controlled by two alleles at a single l
ID: 82860 • Letter: G
Question
Genetics!
Foot webbing color in geese is controlled by two alleles at a single locus and the black allele is completely dominant to the orange allele. You study a population of geese at Karrack Lake, Canada and count 203 geese with orange feet and 473 geese with black feet (676 total). The population is in Hardy-Weinberg equilibrium for this trait. What is the frequency of the black allele in the population? How many of the geese would you expect to be homozvgous black? In the same population of geese, a recessive allele of an X-linked locus produces no mark on the head; the dominant allele produces a white mark. In a population of 330 males, 60 of them have no mark on their head. The population is in Hardy-Weinberg equilibrium for this trait. What is the frequency of the recessive allele? What is the expected frequency of heterozygous females for this trait? A third trait in the same population of geese is feather coloring. Feathers can be brown, black or white. Brown feathers (F^Br) are completely dominant to black feathers (F^BI) and white feathers (f.) Black feathers are completely dominant to white feathers. The allele frequencies for the black allele is 0.28 and the white allele frequency is 0.16. What is the brown allele frequency? What percentage of geese are expected to be white? What percentage of geese are expected to be brown?Explanation / Answer
2) The population is in Hardy-Weinberg equation. Foot webbing color of geese is controlled by two alleles at single locus. Black allele is dominant over the orange allele.
Geese with black feet = 473
Geese with orange feet = 203
Total number of geese = 676
a) Frequency of black allele (p) = 473 / 676 = 0.7
Frequency of orange allele (q) = 203 / 676 = 0.3
b) Homozygous black geese:
frequency of homozygous black allele, p2 = 0.72 = 0.49
Number of homozygous black geese = 676 x 0.49 = 331
c) dominant allele produces white mark
recessive allele produces no mark
60 male geese have dominant allele and 270 have recessive allele
i) Frequency of recessive allele (q) = 270 / 330 = 0.82
ii) Frequency of heterozygous females,
p = 1 - q, = 1 - 0.82, = 0.18
p2 + 2pq + q2 = 1
(0.18)2 + 2(0.18)(0.82) + (0.82)2 = 1
0.03 + 2 (0.15) + 0.67 = 1
pq = 0.15;
therefore, the expected frequency of heterozygous females for this trait is 0.15
d) Extension of hardy-weinberg law to loci with more than two alleles is,
ABC
(p+q+r)2 = p2 (AA) + 2pq (AB) + q2(BB)+ 2pr (AC)+ 2qr(BC) +r2(CC)
black allele frequency (q) = 0.28
white allele frequency (r) = 0.16
i) Brown allele frequency (p) = 1 - (0.28 + 0.16) = 1 - 0.44 = 0.56
ii) percentage of white geese = r2 = (0.16)2 = 0.0256
iii) percentage of brown geese = p2 + 2pq + 2pr [brown color is dominant over black and white]
= (0.56)2 + 2 (0.56) (0.28) + 2 (0.56) (0.16)
= 0.8
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