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Oduvanchick [21]
3 years ago
7

Assume that you work in criminal investigation unit and were called for an incident that involved someone who was shot. You arri

ved to the scene to see a dead person in his 50s and no other evidences rather than a small piece of clothing with blood on it. You collected the blood and went to the lab. The Investigation lead to four suspects that happened to be either close to the place of incident or had problems with the victim. You checked the blood type of that blood and it was (O). When you did blood typing for the four suspects, the results came out as the followings: jacks blood reacted only to A, Sams blood reacted to both A and B, Nathans blood reacted only to B, and Johns blood didnt react to either A or B. Which one of the four is the person of interest (suspect) based on the blood typing reaction?
Biology
1 answer:
ahrayia [7]3 years ago
7 0

Answer: Of the four suspects, JOHN is the person of interest

Explanation:

Since the blood collected from the crime scene was TYPE O, and TYPE O blood DO NOT react to either BLOOD TYPE A or B due to the fact that:

- it do not have antigen A

- it do not have antigen B

JOHN'S BLOOD type is the SAME as that collected from the crime scene. Thus, of the four suspects, JOHN is the person of interest

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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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