Answer:
<em>Liver is the </em><em>l</em><em>argest gland in human body</em><em>.</em>
Answer:
a) the oxidizing agent.
Explanation:
In the citric acid cycle, malate is dehydrogenated into oxaloacetate and the reaction is catalyzed by malate dehydrogenase. The released electrons are accepted by NAD+. So, NAD+ is reduced into NADH. The substances that accept electrons during chemical reactions and are reduced to oxidize the other substances are called oxidizing agents. NAD+ serves as an oxidizing agent as it accepts electrons to oxidize malate into oxaloacetate.
The cross of heterozygous
AaBb as parental phenotype led to formation of 1
6 phenotype of the offspring.This is because this is dihybrid form of inheritance .The probability of having
aabb offspring is
1/16 since there is only one aabb type of offspring phenotype out of
the
16. An illustration of how crossing is done is as per punnet square below
The answer is A. Excretion I hope this helps
Answer:
The fraction of heterozygous individuals in the population is 32/100 that equals 0.32 which is the genotipic proportion for these endividuals.
Explanation:
According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.
In the exposed example, the r-6 allelic frequency is 0,2. This means that if r-6=0.2, then the other allele frequency (R) is=0.8, and the sum of both the allelic frequencies equals one. This is:
p + q = 1
r-6 + R = 1
0.2 + 0.8 = 1
Then, the genotypic proportion for the homozygous individuals RR is 0.8 ² = 0.64
The genotypic proportion for the homozygous individuals r-6r-6 is 0.2² = 0.04
And the genotypic proportion for heterozygous individuals Rr-6 is 2xRxr-6 = 2 x 0.8 x 0.2 = 0.32
