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A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z, arranged in the same order. The distance between X to Y is 10 map unit (mu) and that between X to Z is 25 map. The coefficient of coincidence = 0.88. The percentage of double recombinants in the progeny obtained from the testcross is__________%
Correct answer is '2'. Can you explain this answer?
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A three point test cross was carried out in Drosophila melanogaster in...
Coefficient of coincidence = observed double cross over/Expected double recombinants.
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A three point test cross was carried out in Drosophila melanogaster in...
Calculation of Double Recombinants in Drosophila Melanogaster
Given information:
- Genes: X, Y, and Z arranged in the same order
- Distance between X to Y = 10 mu
- Distance between X to Z = 25 mu
- Coefficient of coincidence = 0.88
To calculate the percentage of double recombinants in the progeny obtained from the testcross, we need to follow the steps below:
Step 1: Determine the expected frequency of double crossovers
- The coefficient of coincidence (C.O.C) = observed double recombinants / expected double recombinants
- C.O.C = 0.88
- The interference (I) = 1 - C.O.C
- I = 0.12
- We know that the expected frequency of double crossovers (DCO) = (recombination frequency between X and Z) x (recombination frequency between Y and Z)
- Recombination frequency between X and Z = 25 mu
- Recombination frequency between Y and Z = (recombination frequency between X and Y) + (recombination frequency between Y and Z) - (recombination frequency between X and Z)
- Substituting the values, we get
- Recombination frequency between Y and Z = 10 mu + (recombination frequency between Y and Z) - 25 mu
- Solving the equation, we get
- Recombination frequency between Y and Z = 15 mu
- DCO = 25 mu x 15 mu = 375 mu^2
Step 2: Determine the expected frequency of non-recombinants, single crossovers, and double crossovers
- The total frequency of non-recombinants, single crossovers, and double crossovers = 1
- The frequency of non-recombinants = (1 - recombination frequency between X and Y) x (1 - recombination frequency between Y and Z) x (1 - recombination frequency between X and Z)
- Substituting the values, we get
- The frequency of non-recombinants = 0.6 x 0.12 x 0.75 = 0.054
- The frequency of single crossovers = (2 x recombination frequency between X and Y) x (1 - recombination frequency between Y and Z) x (1 - recombination frequency between X and Z) + (1 - recombination frequency between X and Y) x (2 x recombination frequency between Y and Z) x (1 - recombination frequency between X and Z) + (1 - recombination frequency between X and Y) x (1 - recombination frequency between Y and Z) x (2 x recombination frequency between X and Z)
- Substituting the values, we get
- The frequency of single crossovers = 2 x 0.4 x 0.12 x 0.75 + 0.6 x 0.15 x 0.75 + 0.6 x 0.12 x 0.25 = 0.396
- The frequency of double crossovers = 1 - frequency of non-recombinants - frequency of single crossovers
- Substituting the values, we get
- The frequency of double
Given information:
- Genes: X, Y, and Z arranged in the same order
- Distance between X to Y = 10 mu
- Distance between X to Z = 25 mu
- Coefficient of coincidence = 0.88
To calculate the percentage of double recombinants in the progeny obtained from the testcross, we need to follow the steps below:
Step 1: Determine the expected frequency of double crossovers
- The coefficient of coincidence (C.O.C) = observed double recombinants / expected double recombinants
- C.O.C = 0.88
- The interference (I) = 1 - C.O.C
- I = 0.12
- We know that the expected frequency of double crossovers (DCO) = (recombination frequency between X and Z) x (recombination frequency between Y and Z)
- Recombination frequency between X and Z = 25 mu
- Recombination frequency between Y and Z = (recombination frequency between X and Y) + (recombination frequency between Y and Z) - (recombination frequency between X and Z)
- Substituting the values, we get
- Recombination frequency between Y and Z = 10 mu + (recombination frequency between Y and Z) - 25 mu
- Solving the equation, we get
- Recombination frequency between Y and Z = 15 mu
- DCO = 25 mu x 15 mu = 375 mu^2
Step 2: Determine the expected frequency of non-recombinants, single crossovers, and double crossovers
- The total frequency of non-recombinants, single crossovers, and double crossovers = 1
- The frequency of non-recombinants = (1 - recombination frequency between X and Y) x (1 - recombination frequency between Y and Z) x (1 - recombination frequency between X and Z)
- Substituting the values, we get
- The frequency of non-recombinants = 0.6 x 0.12 x 0.75 = 0.054
- The frequency of single crossovers = (2 x recombination frequency between X and Y) x (1 - recombination frequency between Y and Z) x (1 - recombination frequency between X and Z) + (1 - recombination frequency between X and Y) x (2 x recombination frequency between Y and Z) x (1 - recombination frequency between X and Z) + (1 - recombination frequency between X and Y) x (1 - recombination frequency between Y and Z) x (2 x recombination frequency between X and Z)
- Substituting the values, we get
- The frequency of single crossovers = 2 x 0.4 x 0.12 x 0.75 + 0.6 x 0.15 x 0.75 + 0.6 x 0.12 x 0.25 = 0.396
- The frequency of double crossovers = 1 - frequency of non-recombinants - frequency of single crossovers
- Substituting the values, we get
- The frequency of double
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A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z, arranged in the same order. The distance between X to Y is 10 map unit (mu) and that between X to Z is 25 map. The coefficient of coincidence = 0.88. The percentage of double recombinants in the progeny obtained from the testcross is__________%Correct answer is '2'. Can you explain this answer?
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A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z, arranged in the same order. The distance between X to Y is 10 map unit (mu) and that between X to Z is 25 map. The coefficient of coincidence = 0.88. The percentage of double recombinants in the progeny obtained from the testcross is__________%Correct answer is '2'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z, arranged in the same order. The distance between X to Y is 10 map unit (mu) and that between X to Z is 25 map. The coefficient of coincidence = 0.88. The percentage of double recombinants in the progeny obtained from the testcross is__________%Correct answer is '2'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z, arranged in the same order. The distance between X to Y is 10 map unit (mu) and that between X to Z is 25 map. The coefficient of coincidence = 0.88. The percentage of double recombinants in the progeny obtained from the testcross is__________%Correct answer is '2'. Can you explain this answer?.
A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z, arranged in the same order. The distance between X to Y is 10 map unit (mu) and that between X to Z is 25 map. The coefficient of coincidence = 0.88. The percentage of double recombinants in the progeny obtained from the testcross is__________%Correct answer is '2'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z, arranged in the same order. The distance between X to Y is 10 map unit (mu) and that between X to Z is 25 map. The coefficient of coincidence = 0.88. The percentage of double recombinants in the progeny obtained from the testcross is__________%Correct answer is '2'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A three point test cross was carried out in Drosophila melanogaster involving three adjacent genes X, Y and Z, arranged in the same order. The distance between X to Y is 10 map unit (mu) and that between X to Z is 25 map. The coefficient of coincidence = 0.88. The percentage of double recombinants in the progeny obtained from the testcross is__________%Correct answer is '2'. Can you explain this answer?.
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