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babymother [125]
3 years ago
13

You have produced antibodies against the GLUT4 glucose transporter and labeled them with a red fluorescent dye. You culture the

cells in the absence of insulin and after a 30-minute incubation period, you fix the cells and treat them with the fluorescent antibody. What do you see when you look at the cells in the fluorescence microscope?
Biology
1 answer:
ycow [4]3 years ago
4 0

The observation of the fluorescent GLUT4 transporter under a microscope of fluorescence confirms the presence of this transporter in the periphery of fibre and in the region of perinucleus.  

Explanation:

The use of red fluorescent dye in labelling of GLUT4 transporter is observed in live cells of mouse. This study showed the predominant presence of GLUT4 in the region of perinucleus and in the fibre periphery. The transporter appeared in small and in large clusters.

The incubation period of 30 minutes and then the glucose ingestion to fix the cell is observed under the view of microscope. This also confirms the constant staining of this protein on the border of the cell of muscle.

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Two autosomal genes, J and K, are 60 map units apart. You perform the following testcross: J K / j k x j k / j k. At what freque
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Answer:

  • J K / j k = 20%
  • j k / j k = 20%
  • J k / j k = 30%
  • j K / j k = 30%                

Explanation:

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.

So, en the exposed example:

  • J and K are autosomal genes
  • J and K are separated by 60 M.U.
  • 60 M.U. means that there is 60% of recombination.

Cross)             J K / j k                    x                  j k / j k

Gametes) JK  Parental                                     jk, jk, jk, jk

                jk   Parental                                  

                Jk   Recombinant                          

                 jK   Recombinant

One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.

1 M.U. -------------- 1% recombination

60 M.U. ------------ 60% recombination

                              30% Jk  +  30% jK

100 M.U. - 60 M.U. = 40 M.U.

40M.U.--------------40 % Parental (Not recombinant)

                            20% JK   +   20% jk

Punnet Square)           JK       jk      Jk      jK

                          jk     JK/jk   jk/jk   Jk/jk   jK/jk

J K / j k = 20%

j k / j k = 20%

J k / j k = 30%

j K / j k = 30%                                

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