Answer:
6,25%
Explanation:
Considering that the couple has a trait of sickle cell anemia, we know that both are heterozygous for the disease (Aa) and therefore can have children with the following genotypes:
Parents: Aa X Aa
Children: AA(A x A), Aa(A x a), Aa (a x A) and aa(a x a)
Knowing that sickle cell anemia only occurs in homozygous individuals, the probability for children to have the disease according to each crossing is:
A x A = 1/4 = 25%
A x a = 1/4 = 25%
a x A = 1/4 = 25%
a x a = 1/4 = 25%
The probability of forming each homozygous child (aa) is 1/4 or 25%. Since they are two children, the probability of both having sickle cell anemia is calculated by multiplying the probability of each, so:
1/4 × 1/4 = 1/16 = 0.0625 = 6.25%
It is concluded that the probability of a heterozygous couple for sickle cell anemia to have two children with the disease is 6.25%.
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Conjugation. I hope this helps!
Is there a image?..
Im confused.
The sequence of bases on RNA that would match or pair with the following DNA sequence: GCA is CGU.
<h3>What is transcription?</h3>
Transcription is the process by which mRNA sequence is produced from a DNA template in the nucleus of a cell.
During transcription process, the following applies:
- Adenine (A) pairs with Uracil (U)
- Guanine (G) pairs with Cytosine (C)
This means that the mRNA sequence that will result from a template of DNA with sequence: GCA is CGU.
Learn more about DNA sequence at: brainly.com/question/3888340
