Use the laws of probability to predict from a trihybrid cross between 2 individuals that are heterozygous for all 3 traits, what expected proportion of the offspring will be: a. homozygous for all three dominant traits b. heterozygous for all three traits c. homozygous recessive for two specific traits and heterozygous for the third

a.) P1 = 0.0156; b.) P2 = 0.125; c.) P3 = 0.09375

Explanation: A trihybrid cross between heterozygous individuals is

AaBbCc x AaBbCc, supposing that alleles A, B and C are any allele.

As the crossing would give a huge variety of genotypes, we can divide each allele and its crossing.

For allele A:

       A        a

A    AA      Aa            

a    Aa       aa

P(AA) = 0.25       P(Aa) = 0.5        P(aa) = 0.25

For allele B:

       B        b

B    BB      Bb

b    Bb       bb

P(BB) = 0.25      P(Bb) = 0.5           P(bb) = 0.25

For allele C:

       C        c

C    CC      Cc

c     Cc      cc

P(CC) = 0.25        P(Cc) = 0.5           P(cc) = 0.25

a.) To be homozygous for all 3 dominant traits, the offspring has to be

AABBCC.  The probability to occur is:

P1 = P(AA)*P(BB)*P(CC)

P1 = 0.25*0.25*0.25

P1 = 0.0156

b.) To be heterozygous, it has to be AaBbCc. So, the proportion will be:

P2 = P(Aa)*P(Bb)*P(Cc)

P2 = 0.5*0.5*0.5

P2 = 0.125

c.) For this, we have to use the 'or' probability, which means we have to sum the proportions, because the genotype of the offspring can be:

aabbCc or Aabbcc or aaBbcc. So it is:

P3 = [P(aa)*P(bb)*P(Cc)] + [P(Aa)*P(bb)*P(cc)] + [P(aa)*P(Bb)*P(cc)]

P3 = (0.25*0.25*0.5)+(0.5*0.25*0.25)+(0.25*0.5*0.25)

P3 = 0.09375

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