Answered: Step Example Perform a cross between a…

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter3: Transmission Of Genes From Generation To Generation

Section: Chapter Questions

Problem 16QP: More Crosses with Pea Plants: The Principle of Independent Assortment Consider the following cross...

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1.) Using the table above as a guide, determine the map distance between the linked genes for eye colour and wing type in the following experiment. You perform the following cross: PpVv × ppvv. You count 1000 offspring in the F1 generation. You find that 406 of the flies have normal eyes and normal wings, 398 have purple eyes and vestigial wings, 96 have normal eyes and vestigial wings, and 100 have purple eyes and normal wings. 2. From previous research findings, you know that the distance between the gene for eye colour and the gene for body colour is 12.2 map units, and the gene for body colour is 7.4 map units away from the gene for wing type. All three genes are on the same chromosome. Draw a chromosome map showing the relative distances between the three linked genes. 3.) You conduct the same cross again, but this time you get an almost exact 1:1 ratio of flies with normal eyes and normal wings to flies with purple eyes and vestigial wings. There are no recombinant types. Provide two explanations that might account for these results.

Step
Example
Perform a cross between a fly that is known to
PpVv × ppvv
be heterozygous for both traits and a fly that is
homozygous recessive for both traits.
Collect a large number of F, flies from the cross. Possible non-recombinant phenotypes:
Determine the number of flies with each of the purple eye, vestigial wing
• normal eye, normal wing
Possible recombinant phenotypes:
• purple eye, normal wing
• normal eye, vestigial wing
possible phenotypes.
Calculate the total number of recombinant F,
phenotypes as a percentage of the total number
of F, flies. This gives you the recombination
frequency and the number of map units
separating the two genes.
number of recombinant types
-x100%
total number of offspring

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