genetics practice problem **Testcross Analysis of Heterozygous Genotype for Genes ABCD**An individual heterozygous for four genes (ABCD) was testcrossed, and the progeny were categorized based on the gametic contribution of the heterozygous parent as follows:- a B C D: 42- A b c d: 43- A B C d: 140- a b c D: 145- a B c D: 6- A b C d: 9- A B c d: 305- a b C D: 310**Task: Construct a Genetic Map**Using the progeny data provided, construct a map of the linked genes. Consider the frequencies of each gametic contribution to identify possible linkage and recombination events among the genes. This will help in visualizing the arrangement and distances between the genes on a chromosome.

Name: genetics practice problem **Testcross Analysis of Heterozygous Genotyp
Uploaded: 2024-03-20T06:42:55.000Z
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Description: genetics practice problem **Testcross Analysis of Heterozygous Genotype for Genes ABCD**An individual heterozygous for four genes (ABCD) was testcrossed, and the progeny were categorized based on the gametic contribution of the heterozygous parent as

The issue raised is a typical illustration of heredity mapping, a method utilized in genetics to decide the relative positions and sequences of genes on chromosomes. This method is established on the guideline of hereditary recombination, which states that novel gene combinations inside the offspring result from the exchange of hereditary fabric between homologous chromosomes.Answer:Based on the data provided, we can construct a map of the linked genes. The process involves identifying the parental and recombinant progeny and calculating the recombination frequencies between the different pairs of genes.The parental progeny are the ones with the highest numbers. In this case, those are "ABcd" and "ab CD". All the other combinations are recombinants.The number of recombinant offspring (for each pair of genes) must be included, divided by the total number of descendants, and multiplied by 100 to get the percentage of recombination frequencies. Recombination frequency (1% recombination frequency = 1 map unit) may be utilized to calculate the distance between genes.Here's how we can calculate the recombination frequencies:For genes A and B: Consider the progeny where only A and B are different from the parental. Those are "Abcd" and "aBCD". The recombination frequency isFor genes B and C: Consider the progeny where only B and C are different from the parental. Those are "ABcD" and "abCD". The recombination frequency isFor genes C and D: Consider the progeny where only C and D are different from the parental. Those are "ABcD" and "abCD". The recombination frequency isHence, the gene order is A-B-C-D and the distances between the genes are: A-B: 1.5, B-C: 28.5, C-D: 8.5.Please note that this is a simplified explanation and actual genetic mapping can be more complex due to factors like crossover interference.The gene order is A-B-C-D and the distances between the genes are: A-B: 1.5, B-C: 28.5, C-D: 8.5.