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olchik [2.2K]
3 years ago
8

The digestive tract lining contains epithelial (outer lining) cells that do not allow food particles to randomly go inside the b

ody except through specific proteins on their cell membrane. Which of the following structure blocks random diffusion of food particles between epithelial cells:A) DesmosomesB) Gap junctionsC) PlasmodesmataD) Tight junctionsE) Cell wall
Biology
1 answer:
Natasha2012 [34]3 years ago
6 0

Answer:

D) Tight junctions

Explanation:

Based on the information provided within the question it can be said that the structure that performs this function are known as tight junctions. This structure has two different functions. The first being to limit the passage of molecules and ions through the space between cells, and the second function is that is blocks the movement of important membrane proteins between the epithelial cells

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Who first suggested the theory of evolution by natural selection?
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Answer:

Charles Darwin

Explanation:

Evolution theory is a shortened form of the term "theory of evolution by natural selection," which was proposed in the 19th century by Charles Darwin and Alfred Russel Wallace.

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Why is it called "hamburger" if there's no ham?​
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What is the role of microfilaments in cell division
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A - the spindle from the kinetochores that attaches to the chrs and pulls them towards the poles is a microfilament.

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Which statement is the most appropriate conclusion for David to write?
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An individual heterozygous for eye color, skin color, and number of eyes mates with an individual who is homozygous recessive fo
sukhopar [10]

Answer:

B) 1 black eyes, green skin, two eyes : 1 black eyes, green skin, one eye : 1 black eyes, white skin, two eyes : 1 orange eyes, green skin, two eyes : 1 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye

Explanation:

Parentals:                     BbGgCc  x  bbggcc

The punette square diagram of the cross of these two parents is attached.

B = black eyes, b = orange eyes;

G = green skin, g = white skin;

C = two eyes, c = one eye

Progeny with BbGgCc genotype → black eyes, green skin, two eyes = 8

Progeny with BbGgcc genotype → black eyes, green skin, one eye = 8

Progeny with BbggCc genotype → black eyes, white skin, two eyes = 8

Progeny with bbGgCc genotype → orange eyes, green skin, two eyes = 8

Progeny with Bbggcc genotype → black eyes, white skin, one eye = 8

Progeny with bbGgcc genotype → orange eyes, green skin, one eye = 8

Progeny with bbggCc genotype → orange eyes, white skin, two eyes = 8

Progeny with bbggcc genotype → orange eyes, white skin, one eye = 8

So, if we will see them in the form of ratio then each of the genotype will be in the ratio of 1 i.e. black eyes, green skin, two eyes: black eyes, green skin, one eye: black eyes, white skin, two eyes: orange eyes, green skin, two eyes: black eyes, white skin, one eye: orange eyes, green skin, one eye: orange eyes, white skin, two eyes : orange eyes, white skin, one eye = 1:1:1:1:1:1:1:1.

So, in view of above option B is most appropriate.

Note: In a trihybrid cross, a total of 64 progeny types may be produced.

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