Gin pea plants, the allele for yellow seeds (y) is dominant over the one for green seeds (y) and the allele for round seeds (r) is dominant over the one for wrinkled seeds (r). a homozygous yellow, round seed is crossed with a homozygous green, wrinkled seed. the offspring of this cross are then crossed. what is the probability that an offspring of the second generation will be yellow and wrinkled? (express your answer as the simplest fraction.)

3/16

Explanation:

The P generation is said to gave homozygous yellow and round seeds; homozygous green and wrinkled seeds. Homozygous means that they have the same alleles for the trait.

For the P generation the cross would be:

YYRR x yyrr

This means that all offsprings in the first generation (F1) would have a genotype of:

YyRr

16/16

(Look at the Punnett attached to see this cross.)

The cross in the F1 generation would then be:

YyRr x YyRr

Now for yellow and wrinkled seeds to come out, the genotype should have at least one Y allele and a pair of r alleles. So the following genotypes is what you need to look out for:

Yyrr

YYrr

So if you will count it you will find:

Yyrr = 2/16

YYrr = 1/16

Just add them up to get the probability:

3/16

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.