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The following X-linked traits are found in fruit flies: vermilion eyes are reces

ID: 217374 • Letter: T

Question

The following X-linked traits are found in fruit flies: vermilion eyes are recessive to red eyes, miniature wings are recessive to long wings, and sable body is recessive to gray body. A cross was made between wild-type males with red eyes, long wings, and gray bodies to females with vermilion eyes, miniature wings, and sable bodies The heterozygous females from this cross, which had red eyes, long wings, and gray bodies, were then crossed to males with vermilion eyes, miniature wings, and sable bodies. The following outcome was obtained: Males and Females 1320 vermilion eyes, miniature wings, sable body 1346 red eyes, long wings, gray body 102 vermilion eyes, miniature wings, gray body 90 red eyes, long wings, sable body 42 vermilion eyes, long wings, gray body 48 red eyes, miniature wings, sable body 2 vermilion eyes, long wings, sable body 1 red eyes, miniature wings, gray body Calculate the map distance between the three genes.

Explanation / Answer

Answer

According to the question, it is a trihybrid cross

Then,

cross                RRLLGG   X   rrllgg

gametes           RLG                rlg

F1                    RrLlGg

Let us suppose F1 individual is Female

So,

                        RrLlGg     X   rrllgg

Gametes          RLG RLg RlG Rlg rLG rLg rlG rlg      X rlg

Genotype F1       RrLlGg, RrLlgg, RrllGg, Rrllgg, rrLlGg, rrLlgg, rrllGg, rrllgg

F2 genotype ration   27:9:9:9:3:3:3:1

From the genotype construction, we have seen that max are a parental type.

From the outcome of a result, maximum individuals are parental types.

It is defined that map distance between genes is equaled to the percentage crossing over frequency.

From the given data

The distance between R­­­­­­­­­­­­­------L

No of an individual of Single crossover + No of individual of a double crossover X 100

                                                            Total no of individual

= 42+48+2+1 X 100

            2951

=3.15cM

The distance between R------g

= No of an individual of Single crossover + No of individual of a double crossover X 100

                                                            Total no of individual

= 102+90+2+1 X 100

            2951

= 6.6cM

Map is R-----------L-------------g