BLACK VESTIGIAL DIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/bl*:vg+/vg+) with a pure-breeding black, vestigial (bl/bl;vg/vg) to produce an F1 generation that is all wild-type (bl+/bl vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (bl+/bl vg+/vg) with tester males, which are black, vestigial (bl/bl vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number 440 Wild-type 394 Black, vestigial 108 Vestigial 135 Black Copy the table into your notes and derive the dihybrid gametes following the example in the previous section. The columns in blue (phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed bl vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete contribution from the offspring genotypes. Note that the two parental gamete types (+ + and bl vg) are the most abundant, as expected. Use the data to calculate the recombination frequency and the genetic map distance between the two genes. Record the map distance in your notes.
BLACK VESTIGIAL DIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/bl*:vg+/vg+) with a pure-breeding black, vestigial (bl/bl;vg/vg) to produce an F1 generation that is all wild-type (bl+/bl vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (bl+/bl vg+/vg) with tester males, which are black, vestigial (bl/bl vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number 440 Wild-type 394 Black, vestigial 108 Vestigial 135 Black Copy the table into your notes and derive the dihybrid gametes following the example in the previous section. The columns in blue (phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed bl vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete contribution from the offspring genotypes. Note that the two parental gamete types (+ + and bl vg) are the most abundant, as expected. Use the data to calculate the recombination frequency and the genetic map distance between the two genes. Record the map distance in your notes.
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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Transcribed Image Text:BLACK VESTIGIAL DIHYBRID TESTCROSS
In the parental generation, you mate a pure-breeding wild-type female (bl+/blt.vg+/vg+) with a pure-breeding black, vestigial (bl/bl vg/vg)
to produce an F1 generation that is all wild-type (bl+/bl vg+/vg). Note that the F1 flies are all dihybrid.
Next, you mate several F1 dihybrid females (bl+/bl vg+/vg) with tester males, which are black, vestigial (bl/bl;vg/vg). The offspring of this
dihybrid testcross are:
Phenotype Genotype Tester Gamete Dihybrid Gamete Number
440
Wild-type
394
108
135
Black, vestigial
Vestigial
Black
Copy the table into your notes and derive the dihybrid gametes following the example in the previous section. The columns in blue
(phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced
from the phenotypes and knowing that the tester contributed bl vg gametes. Finally, you can deduce the dihybrid gametes (green) by
subtracting the tester gamete contribution from the offspring genotypes.
Note that the two parental gamete types (+ + and bl vg) are the most abundant, as expected. Use the data to calculate the recombination
frequency and the genetic map distance between the two genes. Record the map distance in your notes.
Expert Solution
Step 1
If the genes are located on the same chromosome then they are classified as linked genes. In this condition the recombination frequency will be reduced than the normal situation when they are not located on the same chromosome. The recombination frequency is used to calculate the distance between two genes that are located on the same chromosome.
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