If a woman and her husband, who are both carriers, have three children, what is the probability that all three children have the normal phenotype? Express your answer as a fraction using the slash symbol and no spaces (for example, 1/16)

1/64 would be normal

Explanation:

Let the woman phenotype be Yy, since a carrier is usually heterozygous;

Also let the husband phenotype be Yy

Then Yy seperate to yield two alleles "Y" and "y"

So, the cross of the two alleles:

"Y", "y" X "Y", "y"

F1 will be YY, Yy, Yy and yy

From the crossing,

- the normal phenotype are YY,

-the carriers are Yy, Yy

- while yy is recessive.

Since, the probability that one of their offspring is normal is 1/4; for three children, then 1/4 x 1/4 x 1/4 = 1/64

Thus, probability that all three children are normal is 1/64

the correct answer would be 6 feet.

cellular respiration oxidizes food into energy in the form of atps.

glycolysis (6 c hexose) > 2x atps are used > 6c hexose biphosphate > 2 x 3c triphosphate >

2 x 3c pyruvate (2x nadh produced from nad) and (4x atp produced)

pyruvate actively pumped mitochondrial matrix.