A wild type fruit fly heterozygous for the gray body color and normal wings were mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type: 778; black vestigial: 785; black-normal: 158; gray-vestigial: 162.
How many map units separate the genes for body color and wing-type?
A) 8.4
B) 8.6
C) 17
D) 41.3
E) 41.7

17 MU separates the genes for the body color and wing type. Option C.

Explanation:

To know if two genes are linked, we must observe the progeny distribution. If heterozygous individuals, whose genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1.  But if instead of this distribution, we observe a different one, with different phenotypic proportions, we can assume that genes are linked in the double heterozygote parent.

We can recognize which are the recombinant gametes by looking at the phenotypes with lower frequencies in the progeny.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).

The map unit is the distance between a pair of genes for which every 100 meiotic products one of them results in a recombinant one.

We need to know that 1% of recombination frequency = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

In the present example:

Cross: a wild-type fruit fly heterozygous for the gray body color and normal wings with a black fly with vestigial wings.

Let us name the alleles:

                                                            recessive allele g expressing black.

                                                   recessive allele n expressing vestigial wings.

Cross: GgNn   x   ggnn

F1) wild type, GGNN → 778 flies ⇒ Parental

    black vestigial, ggnn → 785 flies ⇒ Parental

    black-normal, ggNn → 158 flies ⇒ Recombinant

    gray-vestigial, Ggnn → 162 flies ⇒ Recombinant

    Total number of flies = 1883

In the present example, the proportion of the F1 has a different distribution than the expected one if genes were not linked. We can assume that genes are linked.

Now, we need to calculate the recombination frequency, P. To do it, we will use the following formula:

P = Recombinant number / Total of individuals

P = 158 + 162 / 1883

P = 320 / 1883

P = 0.1699 ≅ 0.17

To get the genetic distance, we just need to multiply the recombination frequency, P, by 100 and express the result in map units (MU).

P = 0.17

Genetic distance = P x 100 = 0.17 x 100 = 17 MU.

small, big, or no change could occur in the phenotype. it could cause something as simple as a cats ears to curl backwards or something as frighting as death. it could also occur in the protien-coding area and have an effect on that but no effect on the amino acids. what has to occur is a changing in the hox gene or the control gene.