Question

In a laboratory population of Drosophila, all the males are XsY. Among the females, 15% are XiXi, 50% are XiXs, and 35% are XsXs. Assuming random mating, what proportion of male flies in the next generation will be XiY?

Answer 1

Answer:

20 % from the next generation will be male XiY

Step-by-step explanation:

females have are XX always and males are XY, so when cross a woman an a man the probability of have a male is 50 %

XX - XY

⇒XX

⇒XY

⇒XX

⇒XY

P (female) = $\frac{2}{4} =\frac{1}{2}$ = 0.5 = 50 %

P (male) = $\frac{2}{4} =\frac{1}{2}$ = 0.5 = 50 %

Then the probability oh have a male XiY is dependent first oh have a male and second that the male will be XiY

The probability of a dependent probability is the product of the probabilities

P (XiY) = P (male) * P (male wit the character Xi)

P (male wit the character Xi) = P (of be XiY from the amount of males from the cross XiXi with XsY) * P (XiXi) + P (of be XiY from the amount of males from the cross XiXs with XsY) * P (XiXs) + P (of be XiY from the amount of males from the cross XsXs with XsY) * P (XsXs)

1. XiXi - XsY

⇒ XiXs (female)

⇒XiXs (female)

⇒XiY (male)

⇒XiY (male)

From the 2 male, both can be XiY

P (of be XiY from the amount of males from the cross XiXi with XsY) = $\frac{2}{2} =1$

2. XiXs - XsY

⇒ XiXs (female)

⇒XsXs (female)

⇒XiY (male)

⇒XsY (male)

From the 2 male, one can be XiY

P (of be XiY from the amount of males from the cross XiXs with XsY) = $\frac{1}{2}$

3. XsXs - XsY

⇒ XsXs (female)

⇒XsXs (female)

⇒XsY (male)

⇒XsY (male)

From the 2 male,any of them can be XiY

P (of be XiY from the amount of males from the cross XsXs with XsY) = $\frac{0}{2} =0$

P (male wit the character Xi) = P (of be XiY from the amount of males from the cross XiXi with XsY) * P (XiXi) + P (of be XiY from the amount of males from the cross XiXs with XsY) * P (XiXs) + P (of be XiY from the amount of males from the cross XsXs with XsY) * P (XsXs)

P (of be XiY from the amount of males from the cross XiXi with XsY) = 1

P (of be XiY from the amount of males from the cross XiXs with XsY) = $\frac{1}{2}$

P (of be XiY from the amount of males from the cross XsXs with XsY) = 0

P (male wit the character Xi) = $1*\frac{15}{100} +\frac{1}{2} *\frac{50}{100} +0 * \frac{35}{100}$= $\frac{40}{100}=\frac{2}{5}$

P (XiY) = P (male) * P (male wit the character Xi)

P(XiY) = $\frac{1}{2} *\frac{2}{5} =\frac{1}{5} = 0.2$

20 % from the next generation will be male XiY