The scrotum holds the testes.
Answer/Explanation:
For the woman to be blood type B, she must either have 2 B alleles (homozygous, BB) or 1 B allele (heterozygous, BO). We can draw two punnett squares to show each of the potential outcomes when she has children with an AB man.
The two punnett squares are attached. In order to have a child that is type A, she needs to be heterozygous, BO. Even then, there was only a 25% chance that their child would have type A blood. If she were homozygous BB, it would be impossible for her to have a type A child with an AB man.
no podrá obtener estabilidad y no encontrará los recursos a los que está acostumbrado.
Answer and Explanation:
a) Frequencies of A and G lalleles are as follows:
f(A) =( 84+24)/174 = 0.62
f(B) = (42+24)/174 =0.38
b) Expected genotype frequencies:
f(AA) = (0.62) (0.62) = 0.384
f (AG)= 2(0.62) (0.38) =0.471
f(GG) = (0.38) (0.38) = 0.144
c) Genotype Observed Expected O-E (O-E)2 (O-E)2/E
AA 42 33 9 81 2.45
AG 24 41 17 289 7.05
GG 21 13 8 64 4.92
Chi squared = 14.42
The number of degrees of freedom is the number of genotypes minusthe number of alleles= 3-2 =1
The p value is much less than 0.05, therefore we reject the hypothesis that these genotype frequencies may be expected from HArdy Weinberg equilibrum
270/3000=0.09
p + q = 1
So p + 0.09 = 1
p = 0.91
Red eye: 0.91
White eye: 0.09
