Pedigree analysis. The pedigree shows the inheritance of a very rare, completely
ID: 94342 • Letter: P
Question
Pedigree analysis. The pedigree shows the inheritance of a very rare, completely penetrant autosomal recessive disorder. Note that the couple in generation IV are currently expecting their first child (indicated by the "?") What is the probability the child will be affected by the disorder? What is the probability the child will be a carrier (i.e., heterozygous for the causative mutation)? What is the probability the child will be neither affected nor a carrier (i.e., that he/she will be homozygous wild-type)?Explanation / Answer
In generation IV, genotypes of couple are either AA or Aa. AA indicates normal homozygous parent and Aa indicates heterozygous parent.
11. Having affected child (aa), genotypes of parent should be heterozygous (Aa).
Aa X Aa ...............parent
AA Aa Aa aa ........................ F1
So, the probability of affected child (aa) would be – ¼ = 25%.
12. Having heterozygous (Aa) child, genotypes of parent either heterozygous (Aa) or out of two parent one must be normal (AA) and another must be heterozygote (Aa). Here these two cases are discussed below –
Case – 1 Aa X Aa .......................................... Parent
AA Aa Aa aa ................................................ F1
According to this cross, the probability of heterozygous child would be – ½ = 50%
Case – 2 AA X Aa ..........................................Parent
AA Aa ......................................................... F1
According to this case -2, the probability of heterozygous child also – ½ = 50%.
So, we can see here in both cases, the probability of heterozygous child is – ½ = 50%.
13. Having homozygous dominant child (AA), genotypes of parent either heterozygous (Aa) or out of two parent one must be normal (AA) and another must be heterozygote (Aa). Here these two cases are discussed below –
Case – 1 Aa X Aa .......................................... Parent
AA Aa Aa aa ................................................ F1
According to this cross, the probability of homozygous ominant child would be – ¼ = 25%.
Case – 2 AA X Aa ..........................................Parent
AA Aa ......................................................... F1
According to this case -2, the probability of homozygous dominant child would be – ½ = 50%.
Thus, we can say that, there are two different probabilities for homozygous dominant child in these two cases.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.