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In a two-point test cross, 36 of the offspring were recombinant types. The remaining 64 offspring were parental types. How many map units separate the two loci

Sagot :

luisvaschetto

Answer:

36

Explanation:

A two-point test-cross is a cross between an individual with a double heterozygote genotype and a homo-zygous recessive individual in order to determine the recombination frequency between two linked genes. In genetics, one map unit (m.u.) can be defined as the measure of the distance (i.e., genetic distance instead of physical distance) between genes for which one (1) product of meiosis in one hundred (100) is recombinant. In this case, 36 of the offspring have the recombinant phenotype, while the remaining 64 offspring are not recombinant, and therefore both genes are separated by 36 mu (64 + 36 = 100 >> 36 mu).

marianaegarciaperedo

The distance between two genes is usually expressed in map units. In the exposed example, 36 MU separates the two loci.

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The map unit is used to express the distance between a pair of genes.

To interpret it, we need to consider that for every 100 meiotic products, one of them results in a recombinant product.

The genetic distance will result from multiplying the recombination frequency by 100 and expressing it in map units (MU).

To calculate the recombination frequency we will use the next formula:

                 P = Recombinant number / Total of individuals.

We know that

  • Recombinant number = 36
  • Parental number = 64
  • Total number of individuals = 36 + 64 = 100 individuals

So now, all we need to do is to replace the terms by their values

P = Recombinant number / Total of individuals

P = 36 / 100

P = 0.36 ⇒ Recombination frequency

Now we need to get the distance between genes, expressed in Map Units.

P = 0.36

Distance in MU = P x 100 = 0.36 x 100 = 36 MU.

So, 36 MU separates the two loci.

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