A genetic disorder at the cellular level
Hope this helps......!
Answer:
The correct answer is - 1/41,493
Explanation:
Let assume the frequency of the two possible same allele genotype (dominant and recessive) in an inbred population is p and q. Then the frequency of heterozygotes (H) is denoted as:
2pq + 2pqF. ( where F is the inbreeding coefficient).
The frequency of the two different hoozygotes in inbred population can be calculated as:
p2 + pqF and q2 + pqF. (Where p and q are the allele frequency of the dominant and recessive phenotype.
Given: Frequency of Alkaptonuria (q 2) = 1:500, 000
=> q = 1/707
p = 706/707 ( Approx values)
solution:
Inbreeding coefficient (F) = 1/64
Therefore,
Frequency of Alkaptonuria in second cousins= q 2 + pqF
= 1/500, 000 + (706/707 x 1/707) x (1/64)
= 1/500, 000 + 1/45, 248
= 1/41,493 (approx)
Answer:
scientific method can help resolve problems logically
They use a certain kind of camera and the pictures aren't just at once they take one full orbit to take the picture
