Question

A person without the allele for sickle cell anemia has children with a person who is a carrier for sickle cell anemia but doesn’t have the disease. What percentage of their children will have sickle cell anemia? What percentage of their children will be carriers for sickle cell anemia? Show your work with a Punnett square. Queen pls I need help ASAP

Answer 1

Sickle cell anemia is an inherited autosomal recessive disease, which means that only the homozygous recessive individuals have the disease, heterozygous individuals are carriers, and homozygous dominant individuals are normal.

Let "N" represent the normal gene, i.e. without the mutation, and "S" represent the gene with the sickle cell mutation.

The genes of the parent without the disease will be NN, and it will produce gametes with the normal allele "N"

The genes of the parent that carries the disease will be NS and will produce gametes with the normal allele "N" and the recessive allele "S"

For both parents, each allele will have a 0.50 probability, this is granted by the random segregation of genes during meiosis.

To construct the Punnet square, you have to write the gametes of one of the parents in the first row and the gametes of the second parent in the first column. Write the corresponding probabilities for each gamete in each cell:

Cross each column with each row to determine the gene combination of the offspring:

When you cross the first column with the first row, the resulting genotypes will be NN, to determine its probability you have to multiply the probability of the individual inheriting each allele:

`P(N)\cdot P(N)=0.5\cdot0.5=0.25`

Write the result in the Punnet square:

Repeat for all combinations:

-First-row + second-column, the genotypes for the offspring will be NN, and its associated probability is equal to the product of the probabilities of each allele:

`P(N)\cdot P(N)=0.5\cdot0.5=0.25`

-Second-row + first-column, the genotypes of the offspring will be NS, and the associated probability is equal to the product of the probabilities of inheriting each allele:

`P(N)\cdot P(S)=0.5\cdot0.5=0.25`

-Second-row + second-column, the genotypes of the offspring will be NS, and, as before, the probability associated will be:

`P(N)\cdot P(S)=0.5\cdot0.5=0.25`

Write the results in the Punnet square:

As you can see, the offspring will have two possible genotypes NN and NS, to determine the percentage for each you have to add the probabilities of the cells that show the same genotype:

`P(NN)=0.25+0.25=0.50`
`P(NS)=0.25+0.25=0.50`

So, 50% of the children will be homozygous dominant (NN) and 50% of the children will be heterozygous (carriers).

As you can see, since there was only one parent that carried the allele S, there is no possibility that their children will carry the homozygous recessive genotype SS. The probability of their children having sickle cell anemia is 0%