Answer:
The order must be K2→K1, since the permanently active K1 allele (K1a) is able to propagate the signal onward even when its upstream activator K2 is inactive (K2i). The reverse order would have resulted in a failure to signal (K1a→K2i), since the permanently active K1a kinase would be attempting to activate a dead K2i kinase.
Explanation:
- You characterize a double mutant cell that contains K2 with type I mutation and K1 with type II
mutation.
- You observe that the response is seen even when no extracellular signal is provided.
- In the normal pathway, i f K1 activat es K2, we expect t his combinat ion of two m utants to show no response with or without ext racell ular signal. This is because no matt er how active K1 i s, it would be unable to act ivate a mutant K2 that i s an activit y defi cient. If we reverse the order, K2 activating K1, the above observati on is valid. Therefore, in the normal signaling pathway, K2 activates K1.
Answer:
Phospholipids are amphipathic molecules. This means that they have a hydrophilic, polar phosphate head and two hydrophobic fatty acid tails. These components of the phospholipids cause them to orientate themselves, so the phosphate head can interact with water and the fatty acid tails can't, hence forming a bilayer
Explanation:
Answer:
A. He has a large buildup of lactate in his muscles.
Explanation:
Ideally, our body cells, via the process of aerobic cellular respiration, need oxygen to break down glucose. However, during rigorous activities or exercises, oxygen can become unavailable for use. In this case, the body cells switch to use another way of cellular respiration, which is FERMENTATION.
This fermentation causes a build up of lactic acid/lactate (its byproduct) in the cells of the muscles being used, which physically causes a burning and aching sensation just as the case of this hiker. The climbing of the steep path is causing causing the fast usage of oxygen in his cells.
Answer:
15.21 %
Explanation:
If we recall the basic formula of Hardy-Weinberg's equilibrium ; we have the following below:
p + q = 1
p² + 2pq + q² = 1
where;
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Given that p= 0.68 and q = 0.39
the percentage of the homozygous recessive genotype (q² ) will be
(0.39)² = 0.1521
= 0.1521 × 100
= 15.21 %
∴ the percentage of the population that has a homozygous recessive genotype = 15.21 %
