Confusion regarding answer to Probability question
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Confusion regarding answer to Probability question
Two Pure Bred Lines of mice, one with Bald Ears and one with Furry ears. (F= Bald, f=furry)
Cross two lines so F1 gen is Ff x Ff (F is dominant)
Cross F2 lines to get 52:16 breakdown of Bald to Furry (your standard 1/4th)
Question reads as follows:
Pick 2 F2 bald mice, cross to get two F3 mice  both bald.
If you were to cross F3 mice and get two progeny, what is the probability that at at least one of these two F4 mice will have furry ears?
I was told this was the answer
Assume F1 = FfxFf
F2 Crosses to get F3
P(Cross) P(Ff)
FFx Ff 1/3 x 2/3 1/2 x 1/2 = 1/18
Ff x FF 2/3 x 1/3 1/2 x 1/2 = 1/18
Ff x Ff 2/3 x 2/3 2/3 x 2/3 = 16/81
For sake of simplicity (1/18 + 1/18 + 16/81) = Z
F3 Cross to get F4
[Z x Z x 1/4] + [Z x Z x 1/4]  {[Z x Z x 1/4] x [Z x Z x 1/4]}
The part i don't really get is the Probability to get a heterozygote P(Ff). Why is multiplied by two fractions instead of one? ie: why is it (1/2 x 1/2) instead of just 1/2 for say the cross FFxFf.
Cross two lines so F1 gen is Ff x Ff (F is dominant)
Cross F2 lines to get 52:16 breakdown of Bald to Furry (your standard 1/4th)
Question reads as follows:
Pick 2 F2 bald mice, cross to get two F3 mice  both bald.
If you were to cross F3 mice and get two progeny, what is the probability that at at least one of these two F4 mice will have furry ears?
I was told this was the answer
Assume F1 = FfxFf
F2 Crosses to get F3
P(Cross) P(Ff)
FFx Ff 1/3 x 2/3 1/2 x 1/2 = 1/18
Ff x FF 2/3 x 1/3 1/2 x 1/2 = 1/18
Ff x Ff 2/3 x 2/3 2/3 x 2/3 = 16/81
For sake of simplicity (1/18 + 1/18 + 16/81) = Z
F3 Cross to get F4
[Z x Z x 1/4] + [Z x Z x 1/4]  {[Z x Z x 1/4] x [Z x Z x 1/4]}
The part i don't really get is the Probability to get a heterozygote P(Ff). Why is multiplied by two fractions instead of one? ie: why is it (1/2 x 1/2) instead of just 1/2 for say the cross FFxFf.
It might be just me, but I don’t understand any of your crosses or calculations.
F0: FF (bold) x ff (furry) (pure bred lines) all Ff (bold) progeny (F1 generation)
F1: Ff x Ff FF (1): Ff (2) : ff (1) OR 3 bold: 1 furry
Now it gets interesting:
You need to pick two F2 bold mice to breed. You have 1/3 probability of picking FF and 2/3 of chance picking Ff mice.
So it’s either FF x FF FF (F3 all bold probability 1/9) OR
Ff x Ff FF (1): Ff (2) : ff (1) (F3 are 3 bold: 1 furry probabilities are FF 1/9, Ff 2/9, ff 1/9) OR
FF x Ff FF (1) : Ff (1) (F3 all bold probabilities are FF 1/9, Ff 1/9)
 To clarify calculations: probability of FF is 1/3 and probability of Ff is 2/3 , thus probability of this combination/cross is 1/3 x 2/3 = 2/9. Multiply that by probability of progeny 2/9 x 1/2 = 1/9…
So probabilities of F3 are (add) FF 1/3: Ff 1/3 : ff 1/9
Now to get furry mice in F4, you need to breed either
ff (1/9) x ff (1/9) ff (1/81) OR
Ff (1/3 x 1/2 [probability of you choosing Ff over FF bold mouse]) x ff (1/9) … ff (1/3 x 1/2 x 1/9 x 1/2 = 1/108) OR
Ff (1/3 x 1/2) x Ff (1/3 x 1/2) … ff (1/36 x 1/4 = 1/144)
So probability of getting ff in F4 is 1/81 + 1/108 + 1/144 = 0.03…
P.S. Someone, please, verify my calculations…
F0: FF (bold) x ff (furry) (pure bred lines) all Ff (bold) progeny (F1 generation)
F1: Ff x Ff FF (1): Ff (2) : ff (1) OR 3 bold: 1 furry
Now it gets interesting:
You need to pick two F2 bold mice to breed. You have 1/3 probability of picking FF and 2/3 of chance picking Ff mice.
So it’s either FF x FF FF (F3 all bold probability 1/9) OR
Ff x Ff FF (1): Ff (2) : ff (1) (F3 are 3 bold: 1 furry probabilities are FF 1/9, Ff 2/9, ff 1/9) OR
FF x Ff FF (1) : Ff (1) (F3 all bold probabilities are FF 1/9, Ff 1/9)
 To clarify calculations: probability of FF is 1/3 and probability of Ff is 2/3 , thus probability of this combination/cross is 1/3 x 2/3 = 2/9. Multiply that by probability of progeny 2/9 x 1/2 = 1/9…
So probabilities of F3 are (add) FF 1/3: Ff 1/3 : ff 1/9
Now to get furry mice in F4, you need to breed either
ff (1/9) x ff (1/9) ff (1/81) OR
Ff (1/3 x 1/2 [probability of you choosing Ff over FF bold mouse]) x ff (1/9) … ff (1/3 x 1/2 x 1/9 x 1/2 = 1/108) OR
Ff (1/3 x 1/2) x Ff (1/3 x 1/2) … ff (1/36 x 1/4 = 1/144)
So probability of getting ff in F4 is 1/81 + 1/108 + 1/144 = 0.03…
P.S. Someone, please, verify my calculations…
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