Drag each label to the correct location on the image.
1 answer:
Answer:
The offsprings of this punnet square will be Dd, Dd, dd and dd in a ratio 1Dd : 1dd
See attached image for punnet square completion
Explanation:
This question involves a cross involving a single gene with alleles D and d where allele D is dominant over allele d i.e. D will mask the phenotypic expression of d in a heterozygous state.
This cross is between a heterozygous parent (Dd) and a homozygous recessive parent (dd). Parent Dd will produce gametes D and d while parent dd will produce gametes d and d.
Using these gametes in a punnet square, four possible offsprings with two distinct genotypes will be produced. The genotypes are: Dd and dd in a ratio 1:1
Hence, the possibility of having a heterozygous dominant offspring is 1/2 while the possibility of having a homozygous recessive offspring is also 1/2.
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Answer:
- If individual III-1 marries an unaffected, non-carrier female, none of their children will have DMD
Explanation:
From the options, the only correct statements is that <em>If individual III-1 marries an unaffected, non-carrier female, none of their children will have DMD</em>
<u>DMD is an X-linked recessive trait, meaning that a male individual is either affected or unaffected (can never be a carrier). III-1 is unaffected and if he marries an unaffected, non-carrier female, it means that none of their children will have DMD.</u>
All the children of II-4 and II-5 will always come out with DMD because both parents are affected.
Individuals I-1 and II-1 are males. A male can never be a carrier in for X-linked traits. He is either affected or unaffected and as shown by the pedigree, both males are unaffected (unshaded).
