Question

. Mendel's dihybrid crosses produced phenotypes in a 9:3:3:1 ratio. We don't always see 9:3:3:1 ratios in dihybrid crosses, but nonetheless these altered ratios are simply modifications of the classic 9:3:3:1 ratio. True b. False Location: 4.4 Gene Interaction Takes Place When Genes at Multiple Loci Determine a Single Phenotype Gene Interaction with Epistasis Table 4.5 18. Yeast mitochondria have a tubular morphology. In a mutagenic screen, two mutants are isolated that have fragmented mitochondria instead of tubular mitochondria. To determine if these two mutations are at the same or different loci, the two mutants are crossed. The progeny all have tubular mitochondria. What do you conclude? The mutations are at different loci. The mutations are at the same locus Not endugh data is presented. a. c. ocation: 4.4 Gene Interaction Occurs When Genes at Multiple Loci Determine a Sing

Answer 1

The statement that all altered dihybrid cross ratios are modifications of the 9:3:3:1 ratio is false due to possible complex gene interactions like linkage and epistasis. In the yeast mitochondria morphology case, if crossing two mutants produces all tubular mitochondria offspring, it indicates the mutations are at different loci.

Step-by-step explanation:

Mendel's dihybrid crosses exhibit a 9:3:3:1 phenotypic ratio under the assumption of no interaction between the genes. However, when genes interact or are linked, this classic ratio can be altered. Gene linkage can cause genes on the same chromosome to be inherited together, often leading to different phenotypic ratios than what Mendel observed. Therefore, the statement that all altered ratios of dihybrid crosses are simply modifications of the 9:3:3:1 ratio is false; these alterations can arise from more complex interactions such as epistasis or linkage.

In the scenario of yeast mitochondria morphology, if crossing two mutants with fragmented mitochondria yields progeny all with tubular mitochondria, it suggests that the mutations are likely at different loci. This is because the presence of a wild-type allele at either locus in the progeny can compensate for the mutant allele at the other locus, resulting in the wild-type phenotype.