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mina [271]
3 years ago
11

You are comparing two peptides translated in two different cells. The peptide has an identical primary structure, but there is m

uch less of the peptide produced in one of the cells. Which of the following is a potential explanation?
a. an alternative polyA addition site was selected in one of the cells

b. the length of the polyA tail is different in the two cells

c. Alternative splicing was used

d. A and B are both correct.
Biology
1 answer:
eimsori [14]3 years ago
6 0

Length of PolyA tail is different in two cells

Explanation:

  • Different lengths of Poly A tails are present for different mRNAs so when we compare two peptides translated in two different cells we have different lengths of PolyA tail
  • PolyA tail is basically a long chain of nucleotides(adenine nucleotides) which are added to messenger RNA(mRNA) during processing of RNA
  • It increases the stability of the molecule
  • It protects the mRNA molecule from undergoing degradation in the cytoplasm
  • It is involved in binding proteins which further initiates translation;introns are removed from pre mRNA before the final mRNA is exported into cytoplasm
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1. Answer: 900x

A typical microscope has two lens: ocular lens which located near the eye, and objective lens which located near the object. The image will undergo magnification of both lens. The total magnification of the object would be the product of both lens multiplication, not the sum. The multiplication will be: 20x * 45x= 900x

total mag= ocular * object

total mag= 20x * 45x

total mag= 900x


2.  Answer: 100x

The total magnification of the object would be the product of both lens multiplication. If the total magnification 1000x, that mean it was the product of ocular lens and objective lens. If ocular lens magnification is 10x, objective lens magnification would be:

total mag= ocular * object

1000x= 10x*object

object=1000x/ 10x=

object=100x

3. Answer: 0.2 mm

The area of the microscope that can be viewed by the eye should be the same. But since the magnification is different, the actual area that it represent will be different. Microscope with bigger magnification will have smaller diameter of the field of view(DFV).  Remember that micrometer(μm) is 1/1000 of millimeter(mm). The DFV would be:500 μm/ ( 1000x/400x)= 200μm= 0.2 mm


4. Answer: 750μm

This question is similar to number 3 question. Remember that 1 millimeter(mm) equal to 1000 micrometer(μm) .

The DFV of a 10x objective lens is 3 mm. Then the DFV of 40x lens would be:

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DFV2= 3mm/ ( 40x/10x)= 0.75 mm= 750μm

5. Answer: 125μm x 37.5μm

The lens DFV of microscope from question 4 is 750μm, so the width and length of the area would be 750μm.

If 6 organisms could fit across the DFV if they were laid end-to-end, the length would be: 750μm/ 6 organism= 125μm

If 20 organism could fit is stacked side by-side, then the width would be: 750μm/ 20 organism= 37.5μm

6. Answer: 6mm

This question is similar to number 3 and number 4 question. Remember that 1 millimeter(mm) equal to 1000 micrometer(μm).

The DFV of a 100x objective lens is 1.5 mm. Then the DFV of 25x lens would be:

DFV2= DFV1/ (mag1/mag2)

DFV2= 1.5mm/ (100x/25x)= 6 mm


7.Using the 100x objective lens from question 6, you estimate 12 organisms could fit across the DFV if they were laid end-to-end and 30 could fit is stacked side-by-side. What is the length and width of this organism (in microns)?  

Answer: 0.5mm x 0.2mm or 500μm x 200μm

The lens DFV of microscope from question 6 is 6 mm, so the width and length of the area would be 6 mm.

If 12 organisms could fit across the DFV if they were laid end-to-end, the length would be: 6mm/ 12 organism=0.5mm

If 30 organism could fit is stacked side by-side, then the width would be: 6mm/ 30 organism= 0.2mm

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