On the Drosophila X chromosome, the dominant allele y⁺ p...

Question

On the Drosophila X chromosome, the dominant allele y⁺ produces gray body color and the recessive allele y produces yellow body. This gene is linked to one controlling full eye shape by a dominant allele lz⁺ and lozenge eye shape with a recessive allele lz. These genes recombine with a frequency of approximately 28%. The Lz gene is linked to gene F controlling bristle form, where the dominant phenotype is long bristles and the recessive one is forked bristles. The Lz and F genes recombine with a frequency of approximately 32%.

Can any cross reveal genetic linkage between gene Y and gene F? Why or why not?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem together in the fruit fly, just soila. The gene capital S for sepia body color is dominant to its a little lower case s for white body color, add another gene locus on the same chromosome. The gene capital B for normal wings is dominant to, it's a little lower case B or vestigial wings. If the two flies that have the same genotype, the ally's capital S and capital V are on one chromosome and the ally's lowercase S and lower case V are on the homologous chromosome are crossed. Which of the following is the expected phenotypic ratio of the offspring. Assuming that these two gene loci do not cross over the answer. Options are a 1 to 1 B, 1-2-1 C, 3-1 and D, none of these. So this is the type of problem where reading the question carefully is extremely important. So let's review the question information. We know that sepia is dominant to white and normal wings are dominant to vestigial wings. two flies that are crossed have the same genotype, the dominant allys are on one chromosome and the recessive allys are on the homologous chromosome and the cross occurs between these two flies. Now, we would typically expect to do a die hybrid cross. However, and it's the important part. The question states that the two gene loci do not cross over. This means the two genes are linked. So this is a cross between linked jeans and therefore, the law of independent assortment does not apply. That means a parent that he that is heterozygous for S and headers, I guess for V Can only produce two gammy types. A gammy with both dominant alis and a gammy with both recessive ails. Now, we can still use a pundit square to determine the correct answer. And the pundit square results of this cross are on the screen. On the top row is one parent's gas which again, one gammy is the dominant alis and one gammy is the recessive alis. And then the same is happening in the first column for the second parent across their four results in offspring with genotype. One is homozygous dominant for both genes. Two offspring are heterozygous for both genes and one offspring is homozygous recessive for both genes. That means three offspring will appear with sepia body color and have normal wings. These offspring are the ones that are highlighted in green and there is one offspring with white body color and vestigial wings and that offspring is highlighted in the light purple. That means the phenotypic ratio is 3 - one. Now, in linked genes, genetic variation is decreased and the amount of different, a little combinations and gametes produced is greatly reduced. So, again, our phenotypic ratio is option C 3-1. All right, everyone. I hope you found this helpful and I'll see you soon for the next practice problem.

Key Concepts

Genetic Linkage

Recombination Frequency

Independent Assortment

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