Answer:
Answer:
The correct answer would be 0.589
The Hardy-Weinberg equation is p + q = 1
In addition, the frequencies of p and q in a Hardy-Weinberg population remain the same generations over generations.
The value of q⁸ is given as 0.169
So, the value of q = ⇒ 0.411
Thus, the value of p = 1 - 0.411 ⇒ 0.589.
So, the expected value of p would be 0.589
Answer:
a. Heterozygous individuals may pass on their copy of the disease-causing allele to offspring.
Explanation:
Tay-Sachs, which is a recessive lethal disease ---- Let say the recessive lethal diseases is s
∴ it only results when an individual posses two copies of the diseases-causing allele i.e two copies of the disease will be ss.
Now, when two hetrozygous individuals crossed , it is obvious that each can pass on their copy of the disease-causing allele to the offspring.
Let show an illustration for the above statement.
Let the heterozygous individual be Ts, if Ts cross with another Ts;
we will have:
Ts × Ts
T s
T TT Ts
s Ts ss
the offspring are TT,Ts,Ts,ss
We can now see how the Heterozygous individuals pass on their copy of the disease-causing allele to the offspring (Ts).
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Explanation: Thats all i got i hope it helps if not im sorry im having a bad morning
