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Answer:
Based upon the given pedigree chart, one diseased son ( represented by square ) is obtained in the progeny.
Both daughters ( represented by circles ) are disease-free.
We can solve the parental genotype based on trial and error.
Let us consider the first one - father with AA and mother with Aa.
On crossing, progeny obtained will be AA, AA, AA and Aa. One child is heterozygous, but not diseased, so we can leave this option out.
Next, we have father with aa and mother with aa. As both possess recessive alleles, all the progeny will have the disease. This is also not the case.
The third option has father with Aa and mother with Aa. Both are heterozygous for the disease-causing allele.
On crossing, the progeny obtained will be AA, Aa, Aa and aa. One is disease free, two are heterozygous but not diseased and one is diseased. This is the same distribution as in the given progeny. Clearly, this has to be the answer.
Hope I helped.