Examkin

Soru

Zorluk: Çok zorApplications of Genetics in Medicine and Agriculture

A genetic counselor assesses a couple planning to have children. Both parents are asymptomatic carriers of the sickle-cell allele (HbAHbSHb^A Hb^S). The father has blood group ABAB (IAIBI^A I^B) and the mother has blood group OO (iiii). What is the probability that their first child will be a carrier of the sickle-cell trait and also have a blood group that can safely receive red blood cells from a type AA (IAiI^A i) donor without transfusion agglutination?

  1. 14\frac{1}{4} (or 25%25\%)Cevap
  2. B
    12\frac{1}{2} (or 50%50\%)
  3. C
    18\frac{1}{8} (or 12.5%12.5\%)
  4. D
    38\frac{3}{8} (or 37.5%37.5\%)

Cevap

The probability that the child is both a sickle-cell carrier and compatible with type A donor blood is 14\frac{1}{4} (or 25%25\%).
Crossing two sickle-cell carriers (HbAHbS×HbAHbSHb^A Hb^S \times Hb^A Hb^S) gives a 12\frac{1}{2} probability of producing a carrier child (HbAHbSHb^A Hb^S). Crossing a parent of blood type ABAB (IAIBI^A I^B) with a parent of type OO (iiii) yields offspring with blood types AA (IAiI^A i) or BB (IBiI^B i), each with a probability of 12\frac{1}{2}. For red blood cell transfusion from a type AA donor, only type AA offspring can receive the blood safely because type BB offspring possess anti-A antibodies that would cause agglutination. Since the two gene loci assort independently, the overall combined probability is 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} (or 25%25\%).

Adım Adım Çözüm

1
Determine the probability of inheriting the sickle-cell carrier genotype (HbAHbSHb^A Hb^S).
Crossing two carrier parents (HbAHbS×HbAHbSHb^A Hb^S \times Hb^A Hb^S) produces genotypes 1 HbAHbA:2 HbAHbS:1 HbSHbS1\, Hb^A Hb^A : 2\, Hb^A Hb^S : 1\, Hb^S Hb^S. The probability of a carrier offspring (HbAHbSHb^A Hb^S) is 24=12\frac{2}{4} = \frac{1}{2}.
Monohybrid inheritance of autosomal recessive trait yields a 1:2:11:2:1 genotypic ratio.
2
Determine the blood group genotypes of the offspring and identify donor compatibility.
Crossing father ABAB (IAIBI^A I^B) with mother OO (iiii) yields 12\frac{1}{2} blood group AA (IAiI^A i) and 12\frac{1}{2} blood group BB (IBiI^B i). Only blood group AA recipients can safely receive type AA red blood cells without antibody-antigen agglutination.
Blood group BB recipients possess anti-A antibodies in their blood plasma, causing hemolysis/agglutination if transfused with type AA cells.
3
Calculate the combined probability of both independent genetic events.
P(Carrier AND Type A)=P(Carrier)×P(Type A)=12×12=14P(\text{Carrier AND Type A}) = P(\text{Carrier}) \times P(\text{Type A}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} (or 25%25\%).
The hemoglobin locus and ABO locus reside on different chromosomes and assort independently according to Mendel's Second Law.

Anahtar Kavram

Application of Mendel's laws of independent assortment to human medical genetic counseling involving disease carriers and ABO blood transfusion compatibility.
Bu soruyu puanla

Konular

Applications of Genetics in Medicine and Agriculture

Bu konudan daha fazla soru