In peppers the following genes are known O oblate genotype o

In peppers the following genes are known: O (oblate, genotype oo produces flattened fruit), gene P(peach, PP produces hairy fruit) and C(cc-compound in florescence; dominant phenotype single flowers) are H triple heterozygote of unknown origin is test-crossed to produce the following results (+ means dominant phenotype) Which genes are linked, if any, which are unlinked For those genes that are linked construct a map, showing cis/trans arrangement of recessive alleles in this trihybrid What would your conclusion about gene P and C if the test - crossed individual was not heterozygous for the gene O?

Solution

Please note that (110+120=) 230 genotypes are parental out of 1000. So, number of recombinants = 1000-230 = 770

Map distance is 77 map units. For linkage to occur, parental types should be greater than recombinants. Here, the number of recombinants is very high. This means that the genes are not linked. They are assorting independently.

The map will be:

o-------p----------------------c

‘p’ gene will be present in middle. This is evident from your results. Also note that ‘p’ and ‘c’ will be farther in comparison to ‘o’ and ‘p’.

For part ‘C’, you have mentioned that the gene is not heterozygous. This means that the gene is homozygous. You have not mentioned whether the gene is homozygous for wild type or for ‘o’ gene.

Suppose it is homozygous for ‘o’ then the entire progeny will be oblate. If it is homozygous for wild type, then the entire progeny will be wild type, without any oblate phenotype. This is because, linkage is not there.