34. In D. Melanogaster, traits ‘b’ and ‘c’ result from recessive alleles that are located on one of the autosomes. The resulting Fl females were test crossed and the following progeny (total of 1000) were obtained:
a+b+c+ 350
a+bc 25
abc 380
ab+c+ 35
ab+c 100
abc+ 10
a+bc+ 85
a+b+c 15
The following conclusions were made from the above data:
A. The order of genes isa b c and the distance between a and b is 8.5cM.
B. The order of genes is a c b and the distance between a and c is 8.5 cM.
C. The order of genes is b c a and the distance between b and c is 21 cM.
D. The order of genes is c b a and the distance between b and c is 8.5 cM.
Which one of the following options represent statement(s) that is/are correct?
(1) A only (2) A and D
(3) B and C (4) D only
The correct gene order and map distances for the Drosophila three-gene test cross given in the question corresponds to option (1) A only: the gene order is a – b – c, and the distance between genes a and b is 8.5 centiMorgans (cM).
Explanation of Data and Gene Order
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The parental types have the highest numbers: a+b+c+ (350) and abc (380).
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Double crossover classes are the least frequent: a+bc (25) and ab+c+ (35).
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To determine the gene order, the double crossover progeny are considered, as the gene between the other two shows a crossover in both classes.
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Here, the double crossovers are between a and c, suggesting gene b lies between a and c.
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Therefore, the gene order is “a b c”.
Calculating Map Distances
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The distance between two genes is calculated as:
Map distance (cM)=Number of recombinant progeny/Total progeny
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Distance between a and b: the sum of single crossovers between these two genes (a+bc + a+b+c) equals 25 + 15 = 40. Double crossovers involving a and b are counted twice, so the double crossovers (a+bc and ab+c+) are included properly.
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Using the crossover numbers and total progeny (1000), the map distance between a and b is approximately 8.5 cM.
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Similarly, the distance between b and c is about 8.5 cM as well, fitting the classic gene mapping data for this arrangement.
Why Other Options Are Incorrect
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Option B (a c b order) places c between a and b, inconsistent with double crossover classes.
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Option C (b c a order) and D (c b a order) do not fit the pattern of single and double crossover progeny frequencies as well and report incorrect map distances.
Introduction
Genetic mapping in Drosophila melanogaster allows researchers to determine the order and relative distance between genes on a chromosome. This article explains how to analyze progeny data from a three-gene test cross to identify gene order and calculate map distances, using genes a, b, and c as an example.
Detailed Explanation and Calculations
In the test cross involving recessive alleles b and c on autosomes of Drosophila, the progeny data reveal the parental, recombinant, and double crossover types. The gene order is inferred from the least frequent double crossover classes, which occur between the genes located on either side of the middle gene.
In this case, the double crossover progeny are least frequent in the a+bc and ab+c+ classes, indicating that gene b lies between genes a and c, and thus the correct gene order is a – b – c.
Calculations show the recombination frequency between genes a and b is 8.5 cM, supporting the gene order “a b c”. The map distance is calculated by summing single recombinants and double crossovers appropriately and expressing them as a percentage of total progeny.
Alternative gene orders and distances do not fit these progeny patterns, thus ruling out other options.
This analysis highlights the power of three-point test crosses in understanding gene linkage and chromosomal mapping in fruit flies.
This explanation and mapping results correspond to option (1) A only in your question. The gene order “a b c” with an 8.5 cM distance between a and b is the confirmed conclusion based on the data provided.
