A man and his wife are both heterozygous for the sickle cell trait…
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
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)
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:
- A is the right answer. ¾ x 100 = 75%
- B is incorrect.
- C is incorrect.
- 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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