my professor is using the formula in the upper left to solve this. He has a very
ID: 183330 • Letter: M
Question
my professor is using the formula in the upper left to solve this. He has a very thick accent and kinda steam rolled through this problem. I'm not sure how he's getting any of the numbers. please help.
n! (psq"utation-Binomial Theorem P= s! t! In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (at). At an independently assorting locus, an allele for tan coat color (t) is recessive to an allele for black coat color (t). A mouse homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F, are intercrossed to produce the F2. In a litter of eight F2 mice, what is the probability that two will have apricot eyes and tan coats? ik ata+ t+t+ x aa tt F1: F2: a+a t+t at t+ 9/16 a+ tt 3/16 aa t+3/16Explanation / Answer
ANSWER: As we know the basic formula of probabilty is p = n! / s!*t! (ps qt)
where, n = no of events
s = n - no. of outcomes expected
t = no. of outcomes expected
P = probability
we have to calculate the data for F2 generation.
where, n = no of progeny; 8
s = n - no. of outcomes expected ( 8-2 = 6)
t = no. of outcomes expected (2)
p = probability of getting atleast one dominant trait
q = no of offspring having both recessive traits (double recessive)
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