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sergiy2304 [10]
2 years ago
4

In an experimental situation, a student researcher inserts an mRNA molecule into a eukaryotic cell after he has removed its 5' c

ap and poly-A tail. Which of the following would you expect him to find?
Biology
1 answer:
scoundrel [369]2 years ago
7 0

Complete question:

Question: In an experimental situation, a student researcher inserts an mRNA molecule into a eukaryotic cell after he has removed its 5' cap and poly-A tail. Which of the following would you expect him to find?

A) The molecule is digested by restriction enzymes in the nucleus.

B) The MRNA could not exit the nucleus to be translated.  

c) The molecule is digested by exonucleases since it is no longer protected at the 5' end.

D) The cell recognizes the absence of the tail and polyadenylates the mRNA.

E) The molecule attaches to a ribosome and is translated, but more slowly.

Answer:

c) The molecule is digested by exonucleases since it is no longer protected at the 5' end.

Explanation:

The primary transcript formed by transcription in eukaryotes is modified by the addition of a structure called 5' cap at its 5' end. It refers to 7-methylguanosine that is linked to the 5' end of the primary transcript. The presence of 7-methylguanosine at the 5' end protects the mature mRNA against digestion by ribonucleases. Therefore, insertion of mRNA without a 5' cap into a eukaryotic cell would result in the breakdown of the inserted mRNA.

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Which process could occur in maple trees to decrease the chromosome number in pollen cells from 52 down to 26?
dlinn [17]

Meiosis

Explanation:

The process in maple trees that decreases the chromosome number in the pollen cells from 52 to 26 is called meiosis.

Meiosis is an important cell division process in which diploid cells are turned into their haploid number.

  • In meiosis the cells divides into half and produces four daughter cells from the diploid 2n cells.
  • The cell divides twice here. It is common in all sexually reproducing cells.
  • It also occurs in all single celled organisms that reproduces sexually.

Learn more:

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7 0
2 years ago
Y-: yellow S-: star yy: black ss: starless
Ede4ka [16]

Answer:

1. heterozygous yellow and star

2. 37

3. 1/8

4. 168

5. 1/4

Explanation:

Given ,

In f1 generation a cross is made between a true breeding black star bellied sneetch mated with a true breeding yellow starless sneetch

yySS x YYss

It is taken as - Y (yellow) is dominant over y (black)

and S (star) is dominant over s (starless)

1. F1 Generation

Genotype of parents yySS X YYss

gametes - yS, yS, Ys, Ys

All 16 offspring will have genotype YySs

phenotype would be heterozygous yellow and star

2. F2 generation cross

YySs X YySs

YS        Ys         yS        ys

YS YYSS YYSs YySS YySy

Ys YYSs YYss YySs Yyss

yS YySS YySs yySS yySs

ys YySs Yyss yySs yyss

Genotype of offspring are –  

YYSS – 1

YYSs – 2

YySS – 2

YySs – 4

YYss- 1

Yyss- 2

yySS – 1

yySs- 1

yyss- 1

2. Out of 16, 2 are black star bellied sneetches . Which means only 1/8 are black star bellied sneetches

So out of 300, 37 are black star bellied sneetches

3. Only 2 out of 16 are true breeding. i.e 1/8

4. 9 out of 16 are yellow star bellied sneetches, so out of 300, 168 are yellow star bellied sneetches

5. 4 out of 16 are true breeding yellow. Thus, ¼ are true breeding

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4. Jimmy likes to work with his friend Joe in 7th grade science class labs. However, he notices that he tends to get lower grade
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Answer:

If Jimmy works better with someone other than Joe, then Joe is a part of the problem of Jimmy getting lower grades.

b responding variable

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If, then statements are used for scientific hypothesis's.

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