Question

An individual with cd genes was crossed with wild type ++. On test crossing ${F_1}$. The progeny was +c105, d115, cd880 and ++900. Distance between cd genes is?
A. 11 map units
B. 5.5 map units
C. 44 map units
D. 88 map units

Answer 1

Hint: Recombination frequencies or distance between genes can be calculated by dividing the number of recombinants by the total number of offsprings and then multiplying it by hundred. The comparison of recombination frequencies can also help to figure out the order of genes on a chromosome.

Complete answer:
The frequency with which a single chromosomal crossover that takes place during the prophase I of meiosis is known as recombination frequency. Its unit is centimorgan (cM) that describes the recombination frequency of $1\% $. The recombination frequency between c and d genes can be calculated by the formula given below:
Recombination frequency (RF)or distance between genes $ = \dfrac{{\operatorname{Re} combinants}}{{Total\,off\,springs}} \times 100$
According to the information provided in the question;
Total number of individuals/offsprings = 2000
Number of recombinants = 200
Distance between genes $ = \dfrac{{220}}{{2000}} \times 100$
Distance between genes $ = \dfrac{{22}}{2}$
Distance between genes = 11 ma units
Therefore, the distance between cd genes is 11 map units.
Hence, the correct answer is option A.

Additional information: A Test cross is defined as an experimental cross of an individual organism of dominant phenotype but unknown genotype and an organism with a homozygous recessive genotype (and phenotype).

Note: Recombination frequency (RF) is not a direct measure of the distance between the genes on chromosomes. It provides an approximation of the physical distance. The pair of genes having larger recombination frequency is likely farther apart, while a pair of genes having smaller recombination frequency are likely closer together.