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butalik [34]
3 years ago
5

If Pepsin and Trypsin had to function together, which PH would produce the highest activity? How is this answer found?

Biology
2 answers:
san4es73 [151]3 years ago
8 0
The proper environments for Pepsin are acidic, particularly those with a pH of 1.5 to 2. Optimal operating pH for Trypsins is of about 8. It seems that the highest level of activity for the two if they were in the same environment would have to be midway between their optimum range - about pH 5.
Nostrana [21]3 years ago
3 0

Answer:

At neutral pH.

Explanation:

Trypsin and pepsin are protease enzymes that help  in breakdown of large protein molecules. These enzymes are known as digestive enzymes as tripsin is produced by pancreas and pepsin is produced by stomach for digestion.

As pepsin acts on proteins in stomach, it works optimally at acidic pH. This is pH of stomach is acidic due to hydrochloric acid, produced by cells of stomach.

On the other hand, trypsin works optimally at basic pH. This is because trypsin acts on protein molecules in small intestine, where pH is slight basic due to sodium bicarbonate ions.

Thus, if pepsin and trypsin have to function together, they will show highest activity at neutral pH (pH between acidic and basic pH).

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A population of 150 individuals has an allele frequency of 0.2 for the dominant allele (H) and a frequency of 0.8 for the recess
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Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
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