Let
• R = red
• r = white
• S = short stem
• s = long stem
Then two heterozygous (RrSs) parents can each contribute one of the following allele pairs:
RS, Rs, rS, rs
so that the genotypes of their offspring have the following distribution:
[tex]\begin{array}{ccccc} & RS & Rs & rS & rs \\ RS & \boxed{RRSS} & \boxed{RRSs} & \boxed{RrSS} & \boxed{RrSs} \\ Rs & \boxed{RRSs} & RRss & \boxed{RrSs} & Rrss \\ rS & \boxed{RrSS} & \boxed{RrSs} & \boxed{rrSS} & \boxed{rrSs} \\ rs & \boxed{RrSs} & Rrss & \boxed{rrSs} & rrss \end{array}[/tex]
The probability of producing offspring with either the SS or Ss genotype is then 12/16 = 3/4.
In other words, there are 4² = 16 possible genotypes among the offspring, and each parent has a 1/2 probability of contributing the S allele. There are 2 cases to consider:
• both parents contribute the S allele with probability (1/2)² = 1/4
• one parent contributes the S allele while the other contributes the s allele, and this can happen in 2 ways with probability 2 × (1/2)² = 1/2
These events are mutually exclusive, so the total probability is 1/4 + 1/2 = 3/4.
