The family tree shows the inheritance of freckles. Which row in the table shows the probability that the next child produced by parents 1 and 2, and the probability that the next child produced by parents 5 and 6 will have freckles?

Answer:

21) D

We first need to figure out whether the freckles are caused by a recessive allele (let us say ‘f’) or a dominant allele (let us say ‘F’).

Let us start with individuals 5 and 6. Assuming that the condition is caused by recessive allele, then both 5 and 6 must be ff. If they are both ff, then their offsprings must also be ff since there is 100% chance of being ff.

       f             f

f      ff            ff

f      ff            ff

Since their offsprings 7 and 8 do not have the freckles, this means that freckles must be caused by a dominant allele.

This means that since freckles can only be caused by the presence of a dominant allele, individuals 2, 4, 7 and 8 who do not have freckles must be homozygous recessive (i.e. ff).

Let us now look at individuals 1 and 2.

Since 1 has freckles, he can either be FF or Ff. If 1 is FF and 2 is ff, then there is a 100% chance of their offsprings having freckles.

        F             F

f      Ff            Ff

f      Ff            Ff

However, one of their offsprings, 4 does not have freckles. So 1 must be heterozygous.

Since 1 is heterozygous and 2 is homozygous recessive,

       F             f

f      Ff            ff

f      Ff            ff

There is a 50% chance (0.5) that a child produced by 1 and 2 will have freckles.

We know now that any offspring of 1 and 2 who has freckles can only be heterozygous (Ff). So 5 must definitely be heterozygous. 6 cannot be FF, as this means that all of their offsprings would have freckles. Therefore, 6 must be heterozygous as well.

        F             f

F     Ff            Ff

f      Ff            ff

There is a 75% chance (0.75) that a child produced by 5 and 6 will have freckles.