In Achondroplasia the following genotypes and phenotypes are observed.
(A, A = Severe (lethal) A, a = Achondroplasia a, a = Normal height) If a male
with achondroplasia and female with achondroplasia had children, what
would be the chance that their children would have severe (lethal)
achondroplasia?
HELP QUIC

1/4 or 25%

Explanation:

The chance that their children would have severe (;ethal) achondroplasia would be 25% or 1/4.

From the illustration:

AA genotype is physically lethal,

Aa genotype conditions achondroplasia, and

aa genotype conditions normal height.

Now, a male whith achondroplasia (Aa) had children with a female with achondroplasia (Aa):

Aa   x   Aa

Progeny = AA, 2Aa, aa

Out of the 4 children produced from the cross, one of them has AA genotype which represents severe (lethal) conditions. Thus, the chance that their children would have severe (lethal) achondroplasia is 1 out of 4 or 25%.

25%

Explanation:

As shown in the question above, individuals with Achondroplasia have the "Aa" alleles, individuals with a severe form of achondroplasia have the dominant homozygous genotype "AA", while individuals who have the "aa" alleles have normal height.

As we know, the offspring of a cross have an allele from each parent. For this reason, to know what the offspring of a couple with achondroplasia would look like, we must make the punnet square, like the one shown in the figure attached below, to know what the offspring's genotype would be like.

As you can see in the figure, the crossing where the two parents have the genotype "Aa", would give rise to an offspring that would have:

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