Biology 2e
2nd Edition
ISBN: 9781947172517
Author: Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher: OpenStax
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Textbook Question
Chapter 13, Problem 7RQ
Which recombination frequency corresponds to perfect linkage and violates the law of independent assortment?
- 0
- 0.25
- 0.50
- 0.75
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Students have asked these similar questions
What would justify the following ratio appearing after phenotyping the outcome of a crossing
trial: 8.9: 2.9: 3.2:1?
Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.
a
b
C
d
Obviously this represents independent assortment based on crossing dihybrid
heterozygotes.
Obviously this represents gene linkage based on test crossing a dihybrid
heterozygote.
Obviously this represents the results of a trihybrid test cross.
Obviously this represents independent assortment based on crossing monohybrid
heterozygotes.
The recombination frequencies (RF) of genes A, B, C, and D are as follows:
Relationship
B-D
C-D
A-D
B-C
A-B
What is the order of the genes?
OBDAC
OCBAD
O CADB
O ADCB
RF
0.27
0.23
0.21
0.04
0.48
In letter B: If the map distance equals the number of recombinant/total of offspring, wouldn't it be 24/806 x 100? Wouldn't we add both recombinants?
Can you explain letter C? I don't grasp that concept well.
And since I'm using my question already, would you be able to answer D.
Thank you!
Chapter 13 Solutions
Biology 2e
Ch. 13 - Figure 13.3 In a test cross for two...Ch. 13 - Figure 13.4 Which of the following statements is...Ch. 13 - Figure 13.6 Which of the following statements...Ch. 13 - X-linked recessive traits in humans (or in...Ch. 13 - The first suggestion that chromosomes may...Ch. 13 - Which recombination frequency corresponds to...Ch. 13 - Which recombination frequency corresponds to...Ch. 13 - Which of the following codes describes position 12...Ch. 13 - In agriculture, polyploid crops (like coffee,...Ch. 13 - Assume a pericentric inversion occurred in one of...
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- Consider the following Test Cross results & chromosome map: Test Cross Progeny 525 A B C 500 A+ B+C+ 456 A+ B C 308 A B+C+ 101 A B C+ 100 A+ B+C 2 A+ B C+ 1 A B+C Total Progeny = 1993 1-Based on the above data calculate the expectednumber of Double crossovers (DCOs)? Round answers to the closest whole number.arrow_forwardIn a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. Why is the recombination frequency for the outside pair of genes not equal to the sum of the recombination between the adjacent gene pairs?arrow_forwardYou have determined that the gene order for three linked genes being studied is CBA. The number of recombinants resulting from crossover between genes A and B alone totals 40 and 42, while the double-crossover progeny total 4 and 6. What is the recombination frequency between genes A and B if the total number of progeny from the cross is 1000? O 0.102% O 0.092% O 10.2% O 0.082% O 8.2% O 1% O 9.2%arrow_forward
- You have determined that the gene order for three linked genes being studied is A B C. The number of recombinants resulting from crossover between genes A and BC are 44 and 48, respectively, while the double-crossover progeny total 4 and 6. What is the recombination frequency between genes A and B if the total progeny from the cross is 1000? Show all of your work.arrow_forwardWhat is the answer for number 4 and how do you calculate this?arrow_forwardBelow are the results from a 3 gene testcross between an XxYyZz plant and an xxyyzz plant: Testcross outcome: XYZ 365 xyz 367 xYz 110 XyZ 105 XYZ 3 Xyz 4 XYZ 25 xyZ 21 1. Determine the order of these genes on the chromosome. Indicate them on the diagram below. 2. Calculate the map distances between the genes (give you answer as map units). Indicate these distances on the diagram above. 3. Determine the coefficient of coincidence. 4. Determine the interference among these genes.arrow_forward
- What would be the gene map for the following genes with known recombination frequencies? D and A is 5.1 cM, A and C is 15.8 cM, A and B is 7.8 cM, C and D is 10.7 cM, and C and B is 8.0 cM.arrow_forwardHemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?arrow_forwardIn a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the interference value for this data set?arrow_forward
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