Answer: I can only list one (sorry): saber-toothed cats.
Explanation: . . .
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%.
Answer:
A mistaken change in DNA sequencing or coding that could be harmful, not noticed or beneficial
Hope this helps!
Answer:
anwser is C
Explanation:
they both have 6 carbon atoms so A is wrong
both are monosacchride so B is wrong
both have same molecular formula which is (C₆H₁₂O₆) so D is wrong
in glucose the anomeric carbon is the first carbon, whereas in fructose, the anomeric carbon is the second carbon. The anomeric carbon is the one containing the carbonyl group (carbonyl group is a functional group composed of a carbon atom double-bonded to an oxygen atom: C=O)
