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A man and his wife are both heterozygous for the sickle cell trait…

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A man and his wife are both heterozygous for the sickle cell trait. The likely percentage of their offspring that will be either carriers or ‘sicklers’ is

A. 75%

B. 50%

C. 25%

D. 100%




Heterozygous = AS

Homozygous = AA or SS


The man = AS

The woman = AS

Mixing the both, we have:

AS + AS = A(AS) + S(AS)

= AA + AS + SA + SS

= AA + 2AS + SS

This means they will both have:

  • One zero carrier of sickle trait. (AA)
  • Two carriers of sickle trait. (AS) And
  • One sickler. (SS)
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So, out of the four children, the number that will be either a sickler or a carrier = 1 + 2 = 3

Hence the percentage will be = ¾ x 100 = 300/4 = 75%

Now for the right answer to the above question:

  1. A is the right answer. ¾ x 100 = 75%
  2. B is incorrect.
  3. C is incorrect.
  4. D is incorrect.


You may please note this/these:

  • They both will theoretically have four children.
  • One of them will have the disorder.
  • Two of them will be carriers of the trait.
  • And one of them will be totally free.

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/ culled from 2020 JAMB-UTME biology past question 25 /

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