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Bezzdna [24]
3 years ago
6

Can anyone help me with this question??

Biology
1 answer:
hoa [83]3 years ago
8 0

Answer:

3 and 4 are both Aa

8 and 9 cannot be h0m0zygous.

Explanation:

Individual 3 and 4 are non-shaded, which means they are either AA or Aa.

They have three children, 9, 10, and 11. 9 and 11 are also non-shaded, meaning they are either AA or Aa.

However, 10 is shaded, which means he is aa.

For 10 to be aa, he must have gotten one a allele from each parent. Therefore, 3 and 4 must both be heterozygous Aa.

8 and 9 are both non-shaded, which means they are either AA or Aa.

They have three children, 13 is non-shaded, so either AA or Aa.

12 and 14 are both shaded, which means they are both aa. They much each have inherited an a allele from each parent.

This means it is impossible for individuals 8 and 9 to be h0m0zygous.

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Boyle's law states that the volume of a given mass of a gas is inversely proportional to its pressure provided that temperature remains constant.

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The volume of the gas expansion is calculated as follows;

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Learn more about Boyle's law here: brainly.com/question/469270

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