Laws of Probability and a Trihybrid Cross

[betterfuture]

Suppose we cross a trihybrid with purple flowers, and yellow round seeds (heterozygous for all three) with a plant with purple flowers and green, wrinkled seeds (heterozygous for flower color but homozygouse recessive for the other two characters).

Using Mendelian genetics, where the cross is PpYyRr x Ppyyrr, what fraction of offspring from this cross would be predicted to exhibit the recessive phenotypes for at least two of the three characters?
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Help? It is asking to use the law of probability but I don't know how to approach this. I did it one way and got the wrong answer.

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[wizzed101]

First of all, the MCAT will never ask questions like this. I've always hated these kind of BS while studying or doing practice test. They are NOT helpful.

At least 2 out of 3 character means there can be
a) 3 recessive phenotypes, in which case:
for P: 0.25
for Y: 0.5
for R: 0.5
in total = 0.25*0.5*0.5 = 0.0625
b) 2 recessive with the exception of
+ P: 0.75*0.5*0.5 =0.1875
+ V or R: 0.25*0.5*0.5 = 0.0625
in total = 0.1875 + 0.0625*2 (for both cases) = 0.3125

combining both gives 0.375.

 

[betterfuture]

Ahh man! That is awesome! How in the world did you do it so fast?! I was straight staring at the book for an 1 hr.

First of all, can you explain part a). From what it looks like, you did a monohybrid cross of each separate gene and figured out the probability of getting a recessive outcome? For the P you got 1/4 of getting pp, for the Y you got 1/2 of getting yy and for R you got 1/2 of getting rr. That's what I am thinking.

For b), I am lost. Can you explain what you did there? Like in steps. Thank you!

 

[wizzed101]

a) Yes. All of these problems are just extensions of monohybrid crosses, and can be solved as such.
b) there are 3 scenarios:
P dominant, R and Y recessive: 0.1875
R dominant, P and Y recessive: 0.0625
Y dominant, P and R recessive: 0.0625

Glad I could help.

 

[betterfuture]

Am still not understanding where the numbers came from.

You said there are 3 scenarios but I don't get from where you derived the decimal numbers.

b) 2 recessive with the exception of
+ P: 0.75*0.5*0.5 =0.1875
+ V or R: 0.25*0.5*0.5 = 0.0625
in total = 0.1875 + 0.0625*2 (for both cases) = 0.3125

combining both gives 0.375.

 

[wizzed101]

It's similar to
0,75 is the probability of getting the dominant phenotype P, 0.5s are probability of getting recessivie R and Y
So for P dominant, R and Y recessive: 0.75*0.5*0.5 =0.1875
etc...

 

[betterfuture]

Thanks!