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kap26 [50]
1 year ago
13

In certain bacteria, an oval shape (O) is dominant over round (o) and thick cell walls (T) are dominant over thin (t). Draw on a

piece of paper a cross between a heterozygous oval, thick cell walled bacteria with a round, thin cell walled bacteria. What are the phenotype of the F1 and F2 offspring? Complete your drawing, answer the question and upload a photograph of your paper.

Biology
1 answer:
seropon [69]1 year ago
6 0

Answer:

Consider the heterozygous oval, thick cell walled bacteria to have the alleles OoTT and the thin cell walled bacteria to have alleles oott. Results will be 50% oval, thick walled bacteria and 50% round, thick walled bacteria. This will be the F1 progeny.

When the oval, thick walled bacteria from the F1 progeny is cross bred with round, thick walled bacteria then 25 percent of the bacteria will be heterozygous oval, thick walled. 25 percent will be heterozygous oval and heterozygous thick walled. 25 percent will be round and thick walled. 25 percent will be round and thin walled.

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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%                                

3 0
1 year ago
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