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SVETLANKA909090 [29]
3 years ago
6

If a person with AB blood (genotype IAIB) produced an offspring with a person who’s IAi, what’s the chance that the offspring wi

ll have A blood? Give your answer as a percentage. Use the table to make a Punnett square to show your work. Can you show me in a chart to break it down to get that answer which is 50%.
Biology
1 answer:
antiseptic1488 [7]3 years ago
7 0

Answer:

50% of getting A blood type

Explanation:

Using a punnet square you have (sorry cant do a square)

 <u>   IA            IB</u>

<u />

<u>IA</u>

<u />

<u />

<u>i</u>

<u />

Then you you add all the possible combinations from both parents genes

              IA               IB

IA          IA IA           IA IB

i            IA i               IB i

You now have a 50% chance of A type blood because the lowercase (i) wont change the blood type outcome because the IA over shines it.

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Answer:

The correct answer is - XDXd and XD Y

Explanation:

Sex-linked disorders are the disorders that are expressed only when there is two copy of the mutated allele or have not masked by the dominant allele of the gene. In DMD, the dominant allele is XD and the mutated allele is Xd. To express it there are the following genotype possible-

in women XdXd, and in men XdY, so these two genotypes are not possible for the parents in this case. If a mutated allele is present with the dominant allele it will be considered as carrier women but diseased will not be found in them.

So the correct genotype would be

mother: XDXd ( a carrier woman)

father : XDY (normal man)

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