First one is C) A phosphate group is added, and the second is D) Kreb's cycle
To be honest, since we live on Earth (assuming you do as well), Earth is the easiest planet to study. We don't know much about the other planets because we've never been to any of them. I believe that the second easiest planet to study is either Uranus or Saturn.
Answer:
0.0139- frequency of heterozygotes in the population
Explanation:
Let q to the power of 2 represents the frequency of the homozygous recessive (aa) = 1/20000 = 0.00005
This, q = √q^2 = √0.00005 = 0.007
Since p + q = 1
p = 1 - 0.007 = 0.993
Using the formula: p^2 + 2pq + q^2
Where 2pq represents the frequency of the heterozygotes, thing we have
2 x 0.007 x 0.993
= 0.0139
<h3><u>Answer;</u></h3>
C. Mature sex cells
<h3><u>Explanation;</u></h3>
- <u><em>Human somatic cells or body cells are diploid such that they contain two sets of 23 chromosomes which gives a total of 23 chromosomes.</em></u>
- Examples <em><u>of somatic cells in the body that are diploid include, epithelial cells, somatic cells, liver cells, blood cells, nerve cells, etc.</u></em>
- However, <u><em>mature sex cells are haploid, which means they have one set of 23 chromosomes which means they have 23 chromosomes.</em></u> Sex cells are the results of meiosis type of cell division in which a diploid cell divides to form four daughter cells that are haploid.
Answer:
The answer is: The lac operon transcription would be reduced.
Explanation:
The lac operon is the machinery that bacteria uses to break down lactose to use it as an energy source when there is no glucose around. The catabolite activator protein (CAP) senses glucose via a molecule called cAMP. If the levels of glucose are high then no cAMP is produced and therefore no cAMP would be able to bind CAP. That would leave CAP inactive and the transcription would occur but at low levels. The same would happen if CAP protein would have a mutation that would leave it inactive.