Monday, August 20, 2012

STPM Biology - chapter 16: Revision essay question 1

1. A woman and a man who are both normal for blood clotting have a haemophiliac son. Haemophilia is a sex-linked recessive disorder.

Answer


A)  What is meant by sex-linked gene?
Sex-linked gene are genes located on the sex chromosomes but are not involved in determining of sex of an individual. Most of this gene are located on the X chromosomes. The allele located on the same sex chromosomes are not separated during meiosis and is inherited.
B)  What is the probability that their second child will be 1) a haemophiliac boy, 2) a daughter who is a carrier for haemophilia?
Both of the probability is 25%.
C)  Draw a simple pedigree diagram to show that the genotypes of the three individual ( the parents and the haemophiliac son). Explain how the pedigree is obtained.



The son is a haemophiliac. He inherit the X chromosome from the mother and the Y chromosomes from the father. The X chromosome carries the recessive alleles for haemophilia. The mother is phenotypically normal but is a heterozygous carrier. The son inherits the haemophlia recessive allele.

D) Explain why a human population will contain more colour-blind individuals than haemophiliacs?
 Under natural conditions, the selection pressure against haemophilia is greater than that against colour-blindness. Haemophiliacs are weaker than colour-blindness and maynot reach sexual maturiry. Some of the haemophiliac gene are therefore lost from population. However, haemophilia can now be treated by gene therapy and many of them live a normal life.


Assignment submitted by L.Y., Kong,  S. H., Lim, S. C., Soo, C. W. Tan and C.C.,Teh 2011/2012 (IBM)

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