Answer:
frequency of allele “A” is .39
frequency of allele “a” is .77
Explanation:
hardy weinberg equation
p^2 + 2pq + q^2 = 1 or (A+a)^2
p^2 = (AA)
2pq = (Aa)
q^2 = (aa)
AA -> 15/100 = .15
Aa -> 25/100 = .25
aa -> 60/100 = .60
if p^2 = AA = .15 then A = √.15 = 0.38729833462 = .39
if q^2 = aa = .60 then a = √.60 = 0.77459666924 = .77
wikipedia quoraAnkitGauba
Dad: TtBb
Mom: ttbb
You have to use distribution for dihybrid crosses. Meaning, the first allele of each trait has a equal chance of being paired with the other allele of the other trait. So for example with Dad, I will number the traits:
T(1)t(2)B(3)b(4)
To set up the possibilities from Dad, it would be 13, 14, 23, 24: TB, Tb, tB, tb. Same idea goes for Mom, except since all alleles are the same, you only need to make one column for Mom, since if you did all 4, the other 3 would just be repeats of the 1.
Cross:
tb
TB TtBb
Tb Ttbb
tB ttBb
tb ttbb
The phenotypic ratio is 1 Tall Brown: 1 Tall Blue: 1 Short Brown: 1 Short Blue
A. 1000 grams. 1000 is divided by ten each time it goes up the trophic scale. since it goes up 2, 3 and 4, the answer is 1000. (1000/10/10/10=1)
