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zvonat [6]
4 years ago
12

Pepsin, an enzyme that is found in the stomach of humans, functions in breaking down proteins. It is denatured as it travels to

the small intestine and can no longer decompose proteins. Which factor most likely causes the enzyme to denature as it moves through the digestive tract?
A. pH

B. substrate concentration

C. enzyme concentration

D. temperature
Biology
1 answer:
AnnZ [28]4 years ago
7 0

Answer:

A. pH

Explanation:

Denaturation of proteins is a protein whereby the protein molecule loses its primary structure and ultimately its physical functioning as a result of factors including pH, temperature or concentration of chemicals. Based on this definition, it is visible that denaturation of proteins can be caused by these three factors: pH, chemical and temperature.

However, in this case of PEPSIN enzyme, which is a proteinous molecule produced in the stomach of humans and functions in breaking down proteins, its denaturation occurs as it moves along the digestive tract to the intestines. This is as a result of the changes in pH in the digestive tract.

The pH of the stomach where pepsin is produced is acidic i.e. <3.0. However, as one moves towards the intestine, the pH increases, which is unfavorable or considered harsh for the pepsin enzyme. Hence, it gets denatured.

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4 0
3 years ago
In a population of 100 individuals, 50 are genotype MM, 40 are MN, and 10 are NN. What are the frequencies of the M and N allele
mrs_skeptik [129]

Answer:

F(M) = 0.7 F(N)=0.3

Explanation:

We need to take into account that individuals NN and MM only N or M aleles respectively while MN gives just half of the allelic contribution for each. With this in mind we have the following probabilities for getting each of the alleles: MM => p(M) = 1.0 and p(N)=0.0

MN => p(M ) = 0.5 and p(N) = 0.5

NN => P(M) = 0.0 and p(N) = 1.0

With this we can calculate the frequencies of the M and N alleles in the population with 100 individuals using the following formulas

For f(M):

f(M)=\frac{1.0 (MM) + 0.5 (MN)}{Total individuals}

f(M)=\frac{1.0(50)+0.5(40)}{100}

f(M) = \frac{50+20}{100}

f(M)=\frac{70}{100}=\frac{7}{10}=0.7

The frequency of the M allele is 0.7

For f(N):

f(N)=\frac{1.0 (NN) + 0.5 (MN)}{Total individuals}

f(N)=\frac{1.0(10)+0.5(40)}{100}

f(N) = \frac{10+20}{100}

f(M)=\frac{30}{100}=\frac{3}{10}=0.3

The frequency of the N allele is 0.3

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