Number __________________ Polydactyly (PD) is an autosomal dominant trait (polyd

Question

Number

__________________

Polydactyly (PD) is an autosomal dominant trait (polydactyly P; wildtype p). Cystic fibrosis (CF) is an Ma autosomal recessive trait (cystic fibrosis f; wildtype F). A PD woman, otherwise normal in phenotype, marries a healthy normal man. Their 4 children are: 1) normal, 2) PD, 3) CF, 4) CF + PD. What is the probability that their 5th child will have AT LEAST ONE of these conditions? Answer to two decimal places. Part What is the probability the child will be PD? Number Part: What is the probability the child will have cystic fibrosis? Number Part IIl: What is the probability the child will have PD AND cystic fibrosis? Number Part IV: What is the overall probability that the child will have at least one of these condidtions?

Explanation / Answer

Since, from the four children, one is normal, one has PD, one has both and one has only CF, the genotypes of both the parents need to be - PpFf.

Now below is the mendelian cross for the their offspring -

Gamete

From

Mother

For the child to have PD, it needs to have the PP** or Pp**, where * is any possibility of CF gene. Therefore, the probability of child having PD = 12/16 = 0.75. Here 12 is the number of cells with PP** or Pp** and 16 is the number of total cells.

For the child to have CF, its genotype needs to **ff, where * is any possibility of PP gene. Therefore, the probability of child having CF = 4/16 = 0.25. Here 4 is the number of cells with **ff and 16 is the number of total cells.

For child to have both PD and CF, its genotype needs to be, PPff or Ppff. Therefore the probability of child having both PD and CF = 3/16 = 0.1875. Here 3 is the number of cells with either PPff genotype of Ppff genotype and 16 is the total number cells.

Now, children that are normal will have the genotype ppFF or ppFf. Probability of normal children = 3/16 = 0.1875, again 3 is the number of cells with either ppFF pr ppFf genotype and 16 is the total number of cells. Now the remaining children would have atleast PD or CF or both. Therefore by the additive property of probability, their probability would be = 1 - 0.1875 = 0.8125, rounding it off to two decimal points, it would be 0.81.

Gamete from Father

Gamete

From

Mother

PF Pf pF pf PF PPFF PPFf PpFF PpFf Pf PPFf PPff PpFf Ppff pF PpFF PpFf ppFF ppFf pf PpFf Ppff ppFf ppff

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