Question

You observe a "parental" generation cross: NNHH times nnhh. The offspring of that cross are called the "F1" generation. F1 individuals are then crossed with the genotype nnhh.(i.e., a "testcross"). Suppose you count 1000 offspring of the testcross and find the following results: 430 offspring with four fingers on small hands 450 offspring with four fingers on large hands 62 offspring with four fingers on large hands 48 offspring with four fingers on large hands 58 offspring with four fingers on small hands From these data, you should conclude that: The two genes are on separate chromosomes The two genes are on the same chromosome, 12 map units apart The two genes are on the same chromosome, 6 map units apart The two genes are on the same chromosome, 88 map units apart The two genes are on the same chromosome, 58 map units apart

Answer 1

The test cross is a Mendelian cross in which F1 heterozygous population is crossed with homozygous recessive population. If proper Mendelian genetics is followed for such crosses, a stable 1:1:1:1 ratio would have been observed. Deviation from this ratio demonstrate genetic linkage between the concerned genes. According to the data in the question, the normal features would have been long hands and five fingers (homozyogus dominant) or short hands and four fingers (homozygous recessive). Appearance of rest of the phenotypic classes represent recombinant classes.

Thus, the linkage frequency can be translated to map unit distances between the two linked genes would be given by:

Map distance between linked genes = (Total number of recombinants * 100 Map units)/Total number of offsprings

Thus, according to the quantitative data, total number of recombinants = 58+62 or 120

Total population = 1000

Thus recombinant frequency or map distance of two genes = 120*100/1000 or 12.

Thus, this calculation suggests that two genes are linked and located 12 map units apart.

Thus option 2 is correct.