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## Sagot :

**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).

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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