Are more likely to be carriers of x-linked traits, and are more likely to express them

Biology

1 answer:

gavmur [86]3 years ago

3 0

Females are more likely to be carriers of X-linked traits, and males are more likely to express them.

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<span>I is dominant, i is recessive. The A's and B's are just show which allele I is. When there is just one dominant allele, it masks the recessive in blood typing. Remember IA and IB are codominant.

O is always ii
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B is IBi (heterozygous) or IBIB (homozygous)
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Remember: You get one allele from each parent!

1. Father must be ii, mother must be ii, so all children must be ii.

2. Father is IAIA (the homozygous one), the mother is IBIB, so the only possibility for the children is IAIB, because you get one allele from the father and one from the mother.

3. Father is IAi, mother is IBi, so the children can be any of the blood types, because they can have all the combinations of genotypes.

4. Father is ii, mother is IAIB. Children can only be IAi or IBi.

5. Father is IAIB, mother is IAIB. Children can be IAIA, IBIB, or IAIB.

Example of Punnett square:
3. Father is type A, heterozygous, mother is type B, heterozygous
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Mother must be IBi (heterozygous)

_______IA ____ i

IB____ IBIA____IBi

i _____ IAi______ii

Sorry, that was difficult on here, hope it's understandable.
The father's alleles run across the top, the mother's are on the side, you follow to where they meet to find the possibilities for the children. IBIA (AB blood type), IBi (B), IAi (A), and ii (O) are the possibilities in this case.

Hope that helps!</span>

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