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Zorluk: OrtaApplications of Genetics in Medicine and Agriculture

A Rhesus-negative (RhRh^-) woman marries a Rhesus-positive (Rh+Rh^+) man whose father was Rhesus-negative (RhRh^-). What is the probability that their second child will inherit the Rhesus-positive trait and be at risk of developing erythroblastosis fetalis?

  1. 50%Cevap
  2. B
    100%
  3. C
    25%
  4. D
    75%

Cevap

The probability that the child will be Rhesus-positive and at risk is 50%.
The RhRh^- mother has genotype rrrr, while the Rh+Rh^+ father has genotype RrRr because he inherited a recessive rr allele from his RhRh^- father. A monohybrid cross between RrRr and rrrr yields 50% RrRr (Rh+Rh^+) and 50% rrrr (RhRh^-). Because erythroblastosis fetalis occurs when an RhRh^- mother bears an Rh+Rh^+ fetus, there is a 50% chance that any child will inherit the Rh+Rh^+ phenotype and be at risk.

Adım Adım Çözüm

1
Determine the genotype of the mother
Since Rhesus negativity is an autosomal recessive trait, the RhRh^- mother must have the genotype rrrr.
Recessive phenotypes only express when homozygous.
2
Determine the genotype of the father
The father is Rh+Rh^+, so he carries at least one dominant allele (RR). Because his father was RhRh^- (rrrr), he must have inherited a recessive allele (rr). Thus, the father's genotype is RrRr.
An individual receives one allele from each parent.
3
Perform a test cross between the mother (rrrr) and father (RrRr)
The cross Rr×rrRr \times rr produces genotypes RrRr (50%) and rrrr (50%).
Punnett square analysis reveals half the offspring will inherit the RR allele from the father and rr from the mother.
4
Calculate the risk of erythroblastosis fetalis
Erythroblastosis fetalis occurs when an RhRh^- mother carries an Rh+Rh^+ fetus. The probability of the fetus being Rh+Rh^+ (RrRr) is 50%.
The risk applies only to Rh+Rh^+ offspring born to sensitized RhRh^- mothers.

Anahtar Kavram

Rhesus factor inheritance and hemolytic disease of the newborn (erythroblastosis fetalis)
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