Video Transcript
The shape of hemoglobin, the
pigment in red blood cells, can demonstrate codominance. The genotypes and phenotypes are
outlined in the table provided. Both parents have the genotype Hb A
Hb S. What is the probability, percent,
that their offspring will have sickle cell anemia?
Dominant alleles usually completely
suppress the expression of recessive alleles in heterozygotes. For example, imagine that we have
two options for pea color, green and yellow. The allele for yellow peas is
dominant, represented as an uppercase Y, and the allele for green peas is recessive,
represented as a lowercase y. A heterozygous individual with both
alleles in their genotype will produce all yellow peas because the green allele is
masked by the dominant yellow allele. This is called complete
dominance.
However, in codominance, one allele
is not completely dominant to the other, so the phenotype produced by the
heterozygous genotype is unique as a result of both alleles expressing
simultaneously. The question states that an example
of codominance is shown in the shape of hemoglobin, the pigment in red blood
cells. In most people, red blood cells are
coded for by an allele that when homozygous in a person’s genotype makes their red
blood cells biconcave discs, which helps maximize oxygen transport.
This genotype, Hb A Hb A, and the
phenotype it produces, is shown in the first column of the table provided by the
question. A different allele coding for
hemoglobin shape can code for red blood cells to look like sickles, as shown in this
image on the right. The red blood cells all become
flatter and are not as efficient at transporting oxygen when this allele is
homozygous in a person’s genotype, resulting in a condition called sickle cell
anemia.
This genotype, Hb S Hb S, and the
phenotype it produces, is shown in the second column of the table provided by the
question. The alleles for normal red blood
cells and sickle cells are codominant to each other, which means that a heterozygous
genotype would result in both kinds of cells in the phenotype.
This genotype, Hb A Hb S, and the
phenotype it produces, the sickle cell trait, is shown in the final column of the
table provided by the question, which tells us that this does not cause anemia under
normal conditions.
The question tells us that both
parents have this heterozygous genotype and so do not have sickle cell anemia
themselves but do carry the sickle cell trait. We are asked to find the
probability that a child of these two heterozygous parents will have sickle cell
anemia. And we can use a Punnett square to
help us determine the potential outcomes of this cross. First, we put the parents’ alleles
in the top row and left-hand column of the Punnett square, respectively, remembering
that these parents are both heterozygous and so each has one allele that codes for
normal red blood cells, Hb A, and one allele that codes for sickle cells, Hb S.
The four empty squares in the
Punnett square represent the potential genotypes of these two parents’ offspring,
and we fill them in by taking one allele from the corresponding square in the top
row and one allele from the corresponding square in the left-hand column. This is what we should see when we
fill in the other squares in the Punnett square.
We are looking for the probability
that the offspring will have sickle cell anemia, which is coded for by the
homozygous genotype Hb S Hb S. And we can see that only one out of
four of the offspring has this genotype. To work out the probability, in
percent, that the offspring produced by this cross will have sickle cell anemia, we
need to multiply one divided by four by 100 percent. This calculation reveals the
correct answer to this question. The probability that these parents’
offspring will have sickle cell anemia is 25 percent.
