Question

Assume for the purposes of this problem that the order of and distances between three linked genes are the following: unc-32 10mu vab-7 20mu dpy-17 What is the probability of obtaining a phenotypic unc-32 vab-7 offspring from a cross between a heterozygous fly(unc-32 vab-7 dpy-17/ ) and the triple recessive nematode

Answer 1

From the distances between genes and their order, it is possible to get recombination frequencies per region (simple and double recombination frequencies). The probability of obtaining a unc-32 vab-7 + offspring is 0.09 = 9%.  

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We are usually asked to get the order of genes in a chromosome. To do so, we need to get the distances between genes. And to get distances between genes, we need to get their recombination frequency.

When three genes are involved, we get the distances by taking the recombination frequency, P, per region.

We talk about regions, R, when we are making reference to the space between two genes.

                Gene 1 -----(R1)----- Gene 2 --------(R2)------ Gene 3

                                    P1                               P2

• P1 = recombination frequency between Gene 1 and Gene 2

• P2 = recombination frequency between Gene 2 and Gene 3

⇒  P1 = (SR + DR) / N

⇒  P2 = (SR + DR)/ N

Where:

• SR is the number of simple recombinants in each region,

• DR is the number of double recombinants in each region, and

• N is the total number of individuals.

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).  

In the exposed problem, we need to do the inverse way. We already have the order and distances between genes, and we need to get the recombination frequency, P.

We know that the order of genes is as follows,

unc-32 ------------- vab-7----------------- dpy-17

     

Distances are:

• Region 1 = 10 MU

• Region 2 = 20 MU

We need to get the probability of obtaining a phenotypic unc-32 vab-7 + offspring from a cross between a heter0zyg0us fly (unc-32 vab-7 dpy-17/ + + +  ) and the triple recessive nematode (unc-32 vab-7 dpy-17 / unc-32 vab-7 dpy-17).

Cross:  

unc-32 vab-7 dpy-17 / + + +     x     unc-32 vab-7 dpy-17 / unc-32 vab-7 dpy-17

To get the genotype unc-32 vab-7  + , a simple recombination, SR, must occur in region 2.

So we need to get the simple recombination frequency in Region 2.

To do so, first we need to clear the following equation

Distance between vab-7 dpy-17 = P2 x 100 = 20 MU

20MU = P2 x 100

P2 = 20  / 100

P2 = 0.2

The recombination frequency in region 2 (P2) is 0.2.

Now, we need to get the simple recombination frequency in this region.

P2 = (SR + DR) / N

P2 = Simple recombinants Frequency + Double recombinants Frequency

• R1 = 10 MU

• R2 = 20 MU

• Double recombinants Frequency = 10% x 20% = 0.1 x 0.2 = 0.02

P2 = Simple recombinants Frequency + Double recombinants Frequency

0.2 = Simple recombinants Frequency + 0.02

Clearing the equation,

Simple recombinants Frequency = 0.2 - 0.02 = 0.18

Simple recombinant frequency in region 2 will produce the following two chromosomes,

• unc-32 vab-7  +

• +   +   dpy-17

So half of this frequency will produce one chromosome, and the other half will produce the other chromosome.

• unc-32 vab-7  +     ⇒   0.09

• +   +   dpy-17          ⇒   0.09

So, the probability of obtaining a phenotypic unc-32 vab-7 + offspring is 0.09 = 9%.

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