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avanturin [10]
3 years ago
7

You are interested in studying a receptor and decide to make a knockout mouse. However, you notice severe developmental defects

that result in embryonic lethality in this receptor knockout. Which developmental process is most likely affected when you disrupt a receptor on the cell surface, disrupting its ability to receive a signal and initiate a transduction pathway?a. apoptosis cell b. proliferation cellc. cell interactions d. cell movement e. cell differentiation
Biology
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

The correct answer is option c. "cell interactions".

Explanation:

Cell-to-cell interactions, or only cells interactions, refers to the mechanisms that cells have to communicate among them. These mechanisms takes place at the surface of the cells by sending and receiving signals in receptors, which later initiate transduction pathways. Cell interactions play a crucial role in the development of embryos, therefore if a knockout mouse has embryonic lethality due to a disruption in a receptor, it is very likely that the development process that is affected is related with cell interactions.

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A regular progression of species replacement is called __________.
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Answer:

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Explanation:

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3 years ago
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Clathrin coated vesicles bud from eukaryotic plasma membrane fragments when adaptor proteins, clathrin and dynamin-GTP are added
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Answer: Clathrin cages assemble, vesicles form but cannot be pinched of but no disassembly occurs so the vesicles remain coated in clathrin.

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Endocytosis is a cellular mechanism that allows the introduction of extracellular material into the cell. Clathrin-coated vesicles act to incorporate different molecules that are recognized by specific proteins located in the clathrin-coated pits. Upon invagination of a portion of the plasma membrane, the material is transported to its final intracellular destination.

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7 0
3 years ago
Which of the following statements is true?A. An isolated system cannot exchange either matter or energy with its surroundings. B
malfutka [58]

Answer:

A

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An isolated system can NOT exchange energy or matter with it's surroundings. This is why it is isolated.

7 0
3 years ago
The domestic cat genome contains 2.9×109 base pairs. The length of linker DNA in mammals is 50 base pairs. Approximately how man
IgorLugansk [536]

Answer:

Number of nucleosomes in 2.9 * 10^9bp is equal to 1.47 * 10^7

Explanation:

For wounding one nucleosome, total length of DNA required is equal to 146 bp

The length of  linker DNA in mammals is equal to 50 bp

Thus , the total length of DNA that confides between two nucleosome is equal to the sum of wounding length of DNA and the linker length

= 146 + 50\\= 196bp

Thus, in 196bp length of DNA, the total number of nucleosomes is equal to 1

Thus, number of nucleosomes in 2.9 * 10^9bp is equal to

\frac{2.9* 10^9}{196} \\1.47 * 10^7

8 0
3 years ago
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