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Lelu [443]
3 years ago
7

C₆H12O₆ name and count the number of each type of atom

Biology
1 answer:
timama [110]3 years ago
6 0
Carbon: 6
hydrogen: 12
oxygen: 6
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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
Dmitry [639]

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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In which calluses did the new gene insert an existing corn gene?
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Many farmers use chemical herbicides to kill weeds and

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If a new organism were discovered which of the following would most likely be used to classify it into appropriate kingdom???
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Which assumption must be correct for a population to be in hardy-weinberg equilibrium for a specific gene? see section 23.1 ( pa
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There are five basic Hardy-Weinberg assumptions:

1. Random Mating -

2. No natural selection - all members of the parental generation survived and contributed equal numbers of gametes to the gene pool, no matter the genotype

3. No genetic drift (random allele frequency changes) - the population is infinitely large.

4. No gene flow - no new alleles were added by immigration or lost through emigration (no migration)

5. No mutation -  There must be mutation equilibrium.


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