Ask question

Ask question

All categories

4 years ago

8

Osmosis occurs through water channel proteins called... ?

1 answer:

aliya0001 [1]4 years ago

3 0

Ribosomes I believe!

Did I help? Please leave a Thank You on my profile, comment on my profile or help me with my current pending Science question!

Take Care,
Zach1

You might be interested in

Which of the following organelles relies on exocytosis to complete its function?

Katyanochek1 [597]

Answer:

The correct answer is option - Golgi apparatus.

Explanation:

Golgi apparatus is the cell organelle that is integral in sorting, modifying, and packaging the macromolecules for secretion from the cell. The Golgi plays an important function in the synthesis of carbohydrate, and proteoglycans.

The Golgi apparatus packaged proteins into vesicles and these vesicles transported to the target site like cell membrane and secreted by the process of exocytosis.

Thus, the correct answer is option - Golgi apparatus.

3 0

4 years ago

Describe the pathophysiology

xeze [42]

The disorder physiologically processes associated with a disease or injury

7 0

3 years ago

Read 2 more answers

What biosafety levels do most introductory microbiology students work with.

DIA [1.3K]

BSL-1 containment. Is normally found in water testing laboratories, in high schools, and in colleges teaching introductory microbiology classes.

7 0

3 years ago

During which era did the supercontinent Pangaea began to break up?

Nonamiya [84]

The answer ma be C hope help

7 0

3 years ago

Read 2 more answers

In humans there is a dominant allele (A) for the absence of moles; while the recessive allele (a) results in the presence of mol

Art [367]

Answer:

a. $(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

b. i. 0.75

ii. 0.000061

iii. 0.012

iv. 0.17

c. 0.67

Explanation:

a. The expansion of the binomial (p + q)7 would be such that:

$(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

b. Both couples are heterozygous:

Aa x Aa

AA Aa Aa aa

Since A is dominant over a,

probability of having mole (aa) = 1/4

probability of not having moles = 3/4

<em>Therefore, the probability of the first child not having moles </em>= 3/4 or 0.75

ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:

$(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

<em>Probability that all of the children will have moles </em>= $p^0q^7$

since p = 3/4 and q = 1/4

$p^0q^7$ = $(3/4)^0(1/4)^7$ = 0.000061

iii. <em>Probability that the first two children will have no moles and the last five will have moles</em> = $21p^2q5$

= $21(3/4)^2(1/4)^5$

= 0.012

iv. <em>Probability that 4 will have no moles and 3 will have moles out of the 7 children</em> = $35p^4q^3$

= $35(3/4)^4(1/4)^3$

= 0.17

c. <em>Probability that the child born without moles is a carrier of the a-allele = probability of heterozygou</em>s.

From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:

= 2/3 or 0.67

8 0

3 years ago