The genes for miniature wings (m) and garnet eyes (g) are approximately 8 map units apart on chromosome 1 in drosophila. phenotypically wild-type females (m + g / mg +) were mated to miniature-winged males with garnet eyes.
1. which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event?
select all that apply.
a. garnet eyes
b. wild type
c. miniature wings
d. miniature wings, garnet eyes
2. if 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes?

The second question is incomplete. The complete question is as follows :

If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes?

__wild type : __ miniature wings : __ garnet eyes : __ miniature wings, garnet eyes

Enter your answer as the number of flies of each phenotype separated by a colon

1. b. wild type, d. miniature wings, garnet eyes

2. 32 : 368 : 368 : 32

Explanation:

1. Two types of offspring can be produced after mating : parental and recombinant. Recombinants are produced due to crossing over of non sister chromatids at the time of gamete formation. Here, mating occurs between m+g/mg+ female and mg/mg male so the offspring would be divided as follows:

m+g/mg : Parental

mg+/mg : Parental

m+g+/mg : Recombinant

mg/mg : Recombinant

m+g+/mg (wild type) and mg/mg (miniature wings, garnet eyes) are the recombinants hence they were produced due to crossover event.

2. The two genes are 8 mu apart and according to linkage concept map distance in mu = recombination frequency

Hence 8% of the offspring will be recombinants and 92% will be parental type.

wild type = 4% = 0.04 * 800 = 32

miniature wings, normal eyes = 46% = 0.46 * 800 = 368

normal wings, garnet eyes = 46% = 0.46 * 800 = 368

miniature wings, garnet eyes  = 4% = 0.04 * 800 = 32

they have sperm in my gf's mouth

explanation:

the index of refraction of the second medium is higher. i just took the test trust me !