Suppose researchers studied the first chromosome of a new species and mapped three genes of interest, indicated as a, b, and c. A three‑factor cross was performed and the number of gametes formed from the F2 generation was recorded in a table. Note that the order of genes in the table is arbitrary.Calculate the recombination frequencies for each given gene pair in map units.?+ ?+ ?+? ? ?? ?+ ?+?+ ? ?? ?+ ??+ ? ?+? ? ?+?+ ?+ ?1731763533111031After completing the gene mapping, the researchers treated individuals to induce inversions in the chromosomes. After treatment, another three‑factor cross was performed and the number of gametes formed from the F2 generation was recorded in a second table. Note that the order of genes in the table is arbitrary.?+ ?+ ?+? ? ?? ?+ ?+?+ ? ?? ?+ ??+ ? ?+? ? ?+?+ ?+ ?1881852426104946Calculate the recombination frequencies for each given gene pair in map units.

Name: Suppose researchers studied the first chromosome of a new species and
Uploaded: 2024-03-20T06:42:55.000Z
Channel: Bartleby
Description: Suppose researchers studied the first chromosome of a new species and mapped three genes of interest, indicated as a, b, and c. A three‑factor cross was performed and the number of gametes formed from the F2 generation was recorded in a table. Note t

The location of the genes on chromosomes is dependent on crossing over. The genes are linearly arranged on the chromosome. The sequence and the relative distances between various genes is represented graphically by recombination frequencies or cross-over values (COV). Recombination frequency is defined as the percentage of crossing over or recombinants. It is used to create a linkage map of chromosome. Recombination frequency or cross-over value = (number of recombinants / total number of offspring) ×100Case 1: From the given number of gametes, we first need to determine the parental genotype. The genotypes found most frequently are the parental genotypes. From the table, it is clear that the ?+ ?+ ?+ and ? ? ? genotypes are the parental genotypes. The double crossover gametes are always present in the lowest frequency. From the table, we can observe that the ? ?+ ? and ?+ ?+ ? are in the lowest frequency. Total number of individuals = 173+176+35+33+11+10+3+1 = 442Percentage of parental type = (173 + 176) / 442 × 100 = 79%Percentage double crossovers = (3+1) / 442 × 100 = 0.9% The single cross-over in locus 1 and locus 2 constitute the gametes ? ?+ ?+ & ?+ ? ? and ? ?+ ? & ?+ ? ?+ respectively. Percentage of single crossover in locus 1 = (35 + 33) / 442 × 100 = 15.4%Percentage of single crossover in locus 2 = (11 + 10) / 442 × 100 = 4.75%Case 2: From the given number of gametes, we first need to determine the parental genotype. The genotypes found most frequently are the parental genotypes. From the table, it is clear that the ?+ ?+ ?+ and ? ? ? genotypes are the parental genotypes. The double crossover gametes are always present in the lowest frequency. From the table, we can observe that the ? ? ?+ and ?+ ? ?+ are in the lowest frequency. Total number of individuals = 188+185+24+26+1+0+49+46 = 519Percentage of parental type = (188 + 185) / 519 × 100 = 72%Percentage double crossovers = 1 / 519 × 100 = 0.2% The single cross-over in locus 1 and locus 2 constitute the gametes ? ?+ ?+ & ?+ ? ? and ? ? ?+ & ?+ ?+ ? respectively. Percentage of single crossover in locus 1 = (24 + 26) / 519 × 100 = 9.6%Percentage of single crossover in locus 2 = (49 + 46) / 519 × 100 = 18.3%