A scientist mates a female fruit fly heterozygous for two traits with a male fly

Question

A scientist mates a female fruit fly heterozygous for two traits with a male fly homozygous recessive for those two traits. The dominant alleles of the traits are abbreviated A and B, the recessive alleles a and b. When they examined the F1 offspring they found the following data: 300 flies with the Ab phenotype, 300 flies with the aB phenotype, 200 flies with the AB phenotype and 200 flies with the ab phenotype. Select the ONE best answer that explains (fits) this data A. The two traits (genetic loci) are genetically unlinked B. One of the traits is sex-linked, the other is autosomally linked (on an autosome). C. The two traits are genetically linked, with a map distance of 40 map units (centiMorgans) The two traits are genetically linked, with a map distance of 60 map units (centiMorgans) E. The two traits are genetically linked, and sex linked (on the X chromosome) with a map distance of 40 map units. The two traits are genetically linked, and sex linked (on the X chromosome) with a map distance of 60 map units. G. Either C or E could be correct. H. Either D or F could be correct.

Explanation / Answer

The cross between a heterozygous individual with that of a homozygous recessive is called a test cross. This cross is also a test cross as the female is heterozygous and the male is homozygous recessive for two traits. Normal test cross for two traits where the genes segregate independently, show 1:1:1:1 ratio. But if the genes are linked i.e. present on the same chromosome, even if crossing over occurs, then also, the result will be different. The parental types will be present at much higher proportion than the recombinants- for ex. the ratio can be 7:1:1:7( 7 for the parental type and 1 for the recombinants).

For the given number of total 1000 fruit flies expected ratio for a test cross would be 250 : 250 : 250 : 250.

Here, the observed ratio of Ab : aB : AB : ab is 300 : 300 : 200 : 200 which is approximately 1:1 :1 :1. So, it is clear that the genes are not linked. They are present on different chromosomes.

Hence, option A is correct.

Then,Option C and D can not be correct.

In the data given for the F1 flies, nothing is mentioned about the sexes. Also, No traits are mentioned. Hence, it can not be said whether the traits are sex- linked or not. So, option B, E and F can not be considered.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.