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3 years ago

All the major organs of the body are formed by week 10 of gestation. Which process makes this possible?

Biology

2 answers:

klemol [59]3 years ago

7 0

Answer: Differentiation

Explanation: The process of differentiation can be defined as the process where a cell changes from one type to another.

It adds specialization to the cell. It occurs during the time of development and this makes the cell more specialized in their structure and function.

This process helps in the formation of various organs during the gestation period.

Snowcat [4.5K]3 years ago

5 0

D. Differentiation
is the right answer because human bodys structure is a multi-layered dermal system with three main layers that, over time, divide to form the multiple organs in the human body.

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4 years ago

In pea plants, smooth (S) peas are dominant over wrinkled (s) peas in the single gene controlling pea texture. A plant with smoo

OlgaM077 [116]

Answer:

Ss and ss

Explanation:

Since the smooth trait (S) is dominant over the wrinkle trait (s), the genotype of the wrinkled pea parent is definitely ss.

Also, some of the progeny had wrinkled pea texture (ss). Each of the 2 wrinkled alleles must have been donated by each of the parent. It thus means that the smooth pea parent is heterozygous for the trait (Ss).

Now, let us look at a cross between the two parents;

Ss x ss

Progeny: Ss, Ss, ss, ss.

Phenotypically, 50% of the progeny are smooth peas while 50% are wrinkled.

Looking at the population of each of the phenotype resulting from the cross. 252:247 is approximately a 50:50 ratio.

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3 0

3 years ago

The first glycolysis rate-limiting enzyme is under various allosteric regulations. This protein is nearly inactive because of in

navik [9.2K]

Answer:

d. ATP; Fructose-2,6-bisphosphate

Explanation:

Phosphofructokinase-1 is the enzyme that catalyzes the formation of fructose-1,6-bisphosphate from fructose-6-phosphate and ATP. The phosphofructokinase step is the first rate-limiting step of glycolysis.

Phosphofructokinase-1 activity is allosterically regulated. Its activity is increased whenever the cell's ATP supply is depleted or when its breakdown products, ADP and AMP accuulates in he cell. However, it is inhibited when the cell is amply supplied with ATP.

The activity of phosphofructokinase-1 is restored by fructose-2,6-bisphosphate, its most potent activator.

From the given options:

a. AMP :::: citrate is wrong because AMP increases the activity of phosphofructokinase-1 and citrate is not its activator but an inhibitor

b. AMP :::: Fru-2,6-P2 is wrong because AMP increases its activity same as fructose-2,6-bisphosphate

c. ATP :::: citrate is wrong because both citrate and ATP are inhibitors

d. ATP :::: Fru-2,6-P2 is correct as explained above

e. All of these is wrong because of the other wrong options above.

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4 years ago