Blood enters the pulmonary vein with close to 100% of the blinding site for oxygen saturated.
Answer:
d.0.48
Explanation:
When a population is in Hardy Weinberg equilibrium the <u>genotypic </u>frequencies are:
freq (AA) = p²
freq (Aa) = 2pq
freq (aa) = q²
<em>p</em> is the frequency of the dominant <em>A</em> allele and <em>q</em> is the frequency of the recessive <em>a</em> allele.
In this population of 100 individuals, 84 martians have the dominant phenotype and 16 have the recessive phenotype.
Therefore:
q²=16/100
q² = 0.16
q=√0.16
q = 0.4
And p+q=1, so:
p = 1 - q
p = 1-0.4
p = 0.6
The frequency of heterozygotes is:
freq (Aa) = 2pq = 2 × 0.4 × 0.6
freq (Aa) = 0.48
Answer:
i think rhe answer should be c but im not sure
Answer:
1. At the end of S phase- 20 pg DNA
2. At the end of G2 phase- 20 Pg DNA
Explanation:
The cell before undergoing M phase undergoes the steps of interphase that is G₁, S and G₂ phase.
During S phase, the process of cell replication takes place which replicates the DNA as a result of which the amount of DNA doubles. This DNA amount is reduced to half during the anaphase stage of M phase.
In the question since the amount of DNA is 10pg therefore the amount will be double during S phase and becomes 20 pg and will remain 20 pg until the DNA is distributed therefore at the end of G₂ phase Will remain the 20 pg.
It looks like a toothless off of how to train your dragon tbh
