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BigorU [14]
3 years ago
10

Imagine a microbial cell that normally carries out aerobic respiration has encountered a low oxygen environment and may die if t

he levels do not increase.
a. Describe the key role that oxygen plays for this microorganism in the aerobic respiration process and what will occur differently without oxygen that could lead to death. b. How would you class this organism in terms of oxygen for growth requirements?
Biology
1 answer:
weeeeeb [17]3 years ago
7 0

Answer:

a. Oxygen performs an essential role in the mentioned microbial cell in a manner that it takes part in the procedure of glycolysis, Krebs cycle, and electron transport chain, which eventually assists in the production of energy from food substrates and this generation of energy helps the cell to survive.  

In the existence of oxygen, sugar gets dissociated through glycolysis to generate pyruvate, which again in the existence of oxygen is transformed into acetyl CoA. This moves into the Krebs cycle and gets dissociated to water and carbon dioxide generating ATP through ETC. This generation of ATP helps the cell to survive.  

In low oxygen surrounding or in the absence of oxygen, some of the aerobic microbes can switch their respiratory pathway and carry on the process of fermentation and anaerobic respiration to produce energy and thrive. However, the mentioned microbial cell, which when it comes in contact with the low oxygen environment cannot carry out fermentation process and would die eventually.  

b. This organism can be classified as obligate aerobes as they always need oxygen and do not possess the tendency to carry out the process of anaerobic respiration or fermentation under the absence of oxygenic environment.  

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If the frequency of an allele is 0.75, how many copies of this allele should be in a population of 300 diploid individuals?
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I have provided the options below:

a. 150

b. 225

c. 300

d. 450

e. 75

Answer:

150

Explanation:

Hardy-Weinberg law states, the allelic frequencies of a gene that is, q and p follow the relationship p^2+q^2+2pq = 1, if the population is in equilibrium.

Therefore p^2+q^2+2pq = 1, can be used to calculate the percentage of the allelic frequencies i.e. p^2 and q^2

Allelic frequency = 0.75.

Therefore, the percentage of the allele in the population would be given by

0.75^2 = 0.5625 = 56%.

And the total number of individuals with this allele in the population:

number of individuals = 56×300÷100

= 168 or

150 nearest figure.

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