Answer:
lungs:
The blood first enters the right atrium.
The blood then flows through the tricuspid valve into the right ventricle.
When the heart beats, the ventricle pushes blood through the pulmonic valve into the pulmonary artery.
The pulmonary artery carries blood to the lungs where it “picks up” oxygen.
It then leaves the lungs to return to the heart through the pulmonary vein.
The blood enters the left atrium.
It drops through the mitral valve into the left ventricle.
The left ventricle then pumps blood through the aortic valve and into the aorta. The aorta is the artery that feeds the rest of the body through a system of blood vessels.
Blood returns to the heart from the body via two large blood vessels called the superior vena cava and the inferior vena cava. This blood carries little oxygen, as it is returning from the body where oxygen was used.
The vena cavas pump blood into the right atrium and the cycle begins all over again.
Answer:
During the S phase at interphase in meiosis I
Explanation:
During the S (DNA synthesis) phase at interphase in meiosis I, DNA replication occurs here where the chromosomes are doubled. This phase does not occur in meiosis II. At the end of meiosis II, the chromosome number becomes halved in the sex cells. The cell just goes on to divide to ensure haploidy of chromosomes in the gametes such that the sperm from the male and egg from female are both haploid. Fertilization brings about diploidy of the zygote itself
Trace a line into a square
Answer:
0.549 is the frequency of the F allele.
0.495 is the frequency of the Ff genotype.
Explanation:
<em>FF </em>or <em>Ff</em> genotypes determine freckles, <em>ff</em> determines lack of freckels.
In this class of 123 students, 98 have freckles (and 123-98= 25 do not).
If the class is in Hardy-Weinberg equilibrium for this trait, then the genotypic frequency of the ff genotype is:
q²= 25/123
q²=0.203
q= 
q= 0.451
<em>q </em>is the frequency of the recessive f allele.
Given <em>p</em> the frequency of the dominant F allele, we know that:
p+q=1, therefore p=1-q
p=0.549 is the frequency of the F allele.
The frequency of the Ff genotype is 2pq. Therefore:
2pq=2×0.549×0.451
2pq=0.495 is the frequency of the Ff genotype.
By row:
1 | 2
2 | 4
2n (same as parent) | n (half that of the parent)
Somatic (form body tissue) | Reproductive (form gametes)
