Answer:
The probability of having a Ab/ab child is 10%
Explanation:
The genes A/a and B/b are linked and 20 m.u. apart.
The parental cross is:
Gametes:
The man only produces 1 type of gametes, so the probability of him producing an ab gamete is 1.
The woman produces 4: two parental (AB and ab) and two recombinant (Ab, aB).
Man: ab
Woman: AB, ab, Ab, aB
The formula to relate genetic distance with recombination frequency is:
Genetic Distance (m.u.)= Recombination Frequency X 100.
Replacing the data in the formula, we have:
20 m.u. / 100 = Recombination Frequency
0.2 = Recombination Frequency
Because the Recombination Frequency is 0.2, the woman will generate recombinant gametes 20% of times, and parental gametes the other 80%. Each recombinant gamete will appear in 10% of the cases, and each parental gamete will appear in 40% of the cases.
The probailities for each possible genotype of the progeny resulting from that cross will be:
Parental: AB/ab 40%
Parental: ab/ab 40%
Recombinant: Ab/ab 10%
Recombinant: aB/ab 10%
Having the distance between genes is useful to calculate the frequency of each gamete when genes are linked. In the exposed example, the probability of having a first child A b/ a b is 10%.
The probability of crossing over events between two genes.
1% of recombination frequency = 1 map unit = 1cm.
The maximum recombination frequency is always 50%.
When the recombination frequency is inferior to 50%, genes are linked.
Genes that express 50% of recombination frequency or more are not linked genes. They might be in different chromosomes or located in the same one, but too far from each other. Genes are located far enough from each other to assort independently.
Cross)
Parentals: A B/ a b x a b/ a b
Woman's gametes:
We know that there are 20 m.u. between genes, meaning that the recombination frequency is 20% = 0.2. The remaining 80% belongs to not recombinant frequencies.
Since there are two parental gametes and two recombinant gametes, each frequency (20% and 80%) must be divided by two to get the gamete frequency.
According to this information, we can assume that,
Frequencies for each gamete
Man's gametes:
The man produces only ab gametes, so these frequencies equal 1.
Cross: Ab (0.1) x ab (1) = Ab/ab (0.1 = 10%)
According to this calculations, the probability of having a first child A b/ a b is 10%.
You will learn more about recombination frequency at
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