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Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event? Select all that apply. Select all that apply. wild type miniature wings garnet eyes miniature wings, garnet eyes SubmitRequest Answer Part B 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 (example: 100:300:100:300).

Answer :

marianaegarciaperedo

Complete question:

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.

Part A

Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event?

Select all that apply.

  1. garnet eyes
  2. wild type
  3. miniature wings
  4. miniature wings, garnet eyes

Part B

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 (example: 100:300:100:300).

Answer:

Part A:

2. wild type (m+g+/mg)

4. miniature wings, garnet eyes (mg/mg)

Part B:

32:368:368:32  

Explanation:

When calculating the recombination frequency, we need to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.  

The map unit is the distance between the pair of genes for which every 100 meiotic products, only one results in a recombinant one.  

So, en the exposed example:

PART A:

First, we need to Identify the gametes for the individual with the genotype m + g / mg + and label parental and recombinants

Gametes:

  • m+ g parental type
  • m g+ parental type
  • m+ g+ recombinant type
  • m g recombinant type

Knowing that the only possible gamete type for the other parental is mg, then we can identify the phenotypic classes of the offspring that were generated as a result of a crossover event. These would be

2. wild type (m+g+/mg)

4. miniature wings, garnet eyes (mg/mg)

PART B:

  • We know that the total number of individuals equals 800.
  • We also know that the distance between genes equals 8 MU.

To calculate the number of the phenotypes among the offsprings,

          8 map units = 8% of recombination in total  

                               = 8/2% m+g+/mg + 8/2% mg/mg  

                                = 4% + 4%

The recombination frequency of m+g+/mg equals 0.04

The recombination frequency of mg/mg equals 0.04.

               

             100% - 8% = 92% of parental in total =

                               = 92/2% of m+ g/mg + 92/2% m g+/mg  

                               = 46% + 46%

The recombination frequency of m+g/mg equals 0.46

The recombination frequency of mg+/mg equals 0.46.

The frequency for each ggenotype is:

  • F (m+g+/mg) = 0.04  
  • F (mg/mg) = 0.04.
  • F (m+g/mg) = 0.46
  • F (mg+/mg) = 0.46.

The number of individuals with each genotype is:

  • m+g+/mg = 0.04 x 800 = 32
  • mg/mg = 0.04 x 800 = 32
  • m+g/mg = 0.46 x 800 = 368
  • m g+/mg = 0.46 x 800 = 368

32 wild type : 368 miniature wings : 368 garnet eyes : 32 miniature wings, garnet eyes

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