Answer:
Option (D).
Explanation:
Recombination frequency nay be defined as the the frequency of crossover of a single chromosome during meiosis. The maxiumum recombination frequency is 50%.
Genes that has recombination frequency nearly equal to 50% are considered as unlinked genes. Unlinked genes cannot undergo the recombination process and may have equal chances to inherit together or can inherit separately in the next generation.
Thus, the correct answer is option (D).
Answer:
Variability was seen as imperfection, so natural selection could only eliminate the inferior and imperfect, not giving rise to new things, new species.
:
The question is incomplete as it does not have the options which are:
A) creatine phosphate.
B) glycolysis.
C) substrate phosphorylation.
D) oxidative phosphorylation.
E) de novo synthesis.
Answer:
D) oxidative phosphorylation.
Explanation:
The ATP is the energy molecule which provides energy to every metabolic process in the organism.
The ATP in humans is produced by a process called cellular respiration where the last phase of the process called electron transport chain produces the highest amount of protein. The electron transport chain is also known as the oxidative phosphorylation as the oxygen is gained and electrons are lost during the phase.
Thus, D) oxidative phosphorylation is correct.
Answer:
Explanation:
Amount of water displaced by the sphere will be equal to the volume of complete sphere.
As the sphere completely sinks to the bottom of the cup.
We have been given;
Diameter of sphere (d) = 4.6 cm
So,
Volume of sphere (V):
By substituting value of r we get:
1 cm³ = 1 mL
V = 51 mL
So,
Water displaced by the sphere = 51 mL
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
