A man and a woman have 24 children. Of the children, 17 have brown
1 answer:
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Answer:
by testcrossing with a homozygous recessive partner
Explanation:
<u>If a pet cockroach exists whose zygosity is unknown, this can be determined by a test cross. A test cross involves crossing an organisms whose zygosity is unknown with a partner that is homozygous recessive for the same trait.</u>
Let us assume that brown body is represented by the allele B, the dominant allele. The homozygous recessive version would be bb.
The genotype of a brown cockroach whose zygosity is not known can be denoted as B_, where '_' can be a 'B' or a 'b'.
When B_ is crossed with bb:
B_ x bb
Progeny
2 Bb
2 _b
The phenotype of Bb would be brown (since B is dominant over b) while the phenotype of _b would depend on the zygosity level of the cockroach.
If the unknown genotype is BB, then _b becomes Bb and the phenotype will be a brown body. This means that all the progeny will appear brown. (<em>see the first attached image for the Punnet's square</em>)
In other word, if the unknown genotype is bb, then _b becomes bb and the phenotype will be a alternate color (non-brown) body. This means that 50% of the progeny will appear brown while the remaining 50% will be in the alternate color. (<em>attached</em>
