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faltersainse [42]
3 years ago
8

As food travels through the digestive system it is exposed to a variety of pH levels. The stomach has a pH of 2 due to the prese

nce of Hydrochloride acid and the smallest intestine has a pH ranging from 7 to 9. HCI converts pepsinogen into pepsin an enzyme that digests proteins in the stomach. Which of the following most likely happens to pepsin as it enters the small intestine.
The pepsin begins to replicate
The pepsin’s activity increases to digest more proteins
The pepsin becomes inactive
The pepsin’s shape changes to engulf large proteins
Biology
1 answer:
True [87]3 years ago
7 0

Answer:

The answer is C. Pepsin becomes inactive as soon as it reaches small intestine.

Explanation:

Enzymes are pH specific. They once work at those specific values and if those are changed they either degrade or become inactive. The pH of stomach is acidic and that of the small intestine is basic. Therefore the enzymes will become inactive when it enters small intestine.

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Answer:

Embryonic period

Explanation:

The time period of the growth of the baby in the womb of the mother of human is 38 weeks or 9 months.

The time period of 38 weeks is divided into three phases related to the events taking place in that period which are: the germinal period, embryonic period and the fetal period.

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  • The embryonic phase begins after the second week and lasts about the eighth week which is marked by the phase of embryonic and differentiation.

Therefore, in the fifth week she is in the embryonic phase of prenatal development.

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Andru [333]

Answer:

1500

Explanation:

Let's assume that the allele for yellow seed color is "Y" and the allele for green seed color is "y". Genotype of pure breeding yellow seeded plant would be "YY" and that of the green seeded plant would be "yy". A cross between YY and yy gives all heterozygous yellow seeded plants (Yy) in F1 progeny. Self pollination of two F1 plants (Yy x Yy) obtains F2 generation in 3 yellow: 1 green ratio.

The total population size of F2 generation = 2000

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Therefore, total number of yellow seeded plants in F2 progeny = 3/4 x 2000= 1500

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