Typically, alleles separate independently. If they're linked, allele separation isn't subject to a normal distribution.
Their parents' phenotypes tell you that each is Rr heterozygous and Ee heterozygous (each Rh+ and E/-: we know that they are not homozygous EE because they each have an unaffected parent).
Therefore, Tom and Terri are both RrEe (RE/++).
They're both:
R E
===========
r e
Typically, each of the four allele groupings would be equally likely (RE, re, Re, eR). Since there's linkage and there's an RF of 4%, however, the parentals are 96% likely, and the recombinants (Re/rE) are only 4% likely:
RE 48% (parental genotype)
re 48% (parental genotype)
Re 2% (recombinant)
rE 2% (recombinant)
1) What is the probability that the first child of Tom and Terri will be Rh- and have elliptocytosis?
Offspring genotype: rrEE or rrEe
rrEE means that both Tom and Terri produced rE gametes (recombinants). Since each recombinant has a 2% likelihood, then .02*.02 = 0.0004
rrEe means that one parent produced an re gamete (parental genotype - 48% chance) and one produced a recombinant (2% chance): .02*.48 = 0.0096
You need to multiply this by 2 since there are two ways to produce this (depending on who produced the recombinant: 0.0096*2 = 0.0192
Now, the total probability of the event occurring is the sum of 0.0192 + 0.0004 =
0.01962) What is the probability that a child of Tom and Terri who is Rh+ will have elliptocytosis?
Offspring genotype:
RREE, RREe, RrEE, RrEe
RREE: .48*48 = .2304
RREe: .48*.02*2 = .0192
RrEE: .48*.02*2 = .0192
RrEe:
4 ways (2 from each parent): RE x re (.48*.48*2) + Re x rE (.02*.02*2) = 0.4616
Sum of these:
0.7304Makes sense.... dominant traits with a low recombination frequency. I attached some frequency charts to help this make a little more sense (I hope). There is probably a more direct way, but this is my 3 AM brute force method.
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