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In dogs, black fur is dominant to white fur. Two heterozygous black dogs are mated. What would be the probability of the following combinations of offspring in a litter of six pups?

a) All with black fur.
b) Four with black fur and two with white fur.
c) At least two of the pups having white fur.
d) The firstborn pup with white fur, and among the remaining five pups, three with black fur and two with white fur.

Respuesta :

obasola1

Answer:

a.p=0.177

b.0.296

c. 0.463

D. 0.066

Explanation:

Since black fur is dominant over white fur, we say

probability of giving back to offsprings with blackfur=75%

whitefur offspring is 25%

All with black fur

P=[tex]^{n} C_{r} B^{r} W^{n-r}[/tex]

P=6C6*.75^6*0.25^(6-6)

P=0.177*1*1

p=0.177

b) Four with black fur and two with white fur

6C4*0.75^4*0.25^2

0.296

cc) At least two of the pups having white fur

if two of the pups have at least white fur

the combinations will be

6C4*0.75^4*0.25^2+6C3*0.75^3*0.25^3+6C2*0.75^2*0.25^4+6C1*0.75^1*0.25^5+6C6*0.75^0*0.25^6

0.296+0.131+0.032+0.0043+0.0002

0.463

d. The firstborn pup with white fur, and among the remaining five pups, three with black fur and two with white fur.

6C5(0.75)^5*0.25^1*5C3*.75^3+0.25^2

p=0.066

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