Question

Another cross in Drosophila involved the recessive, X-linked genes yellow \((y),\) white \((w),\) and \(c u t(c t) .\) A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The \(\mathrm{F}_{1}\) females were wild type for all three traits, while the \(\mathrm{F}_{1}\) males expressed the yellow-body and white- eye traits. The cross was carried to an \(\mathrm{F}_{2}\) progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed. (a) Diagram the genotypes of the \(\mathrm{F}_{1}\) parents. (b) Construct a map, assuming that white is at locus 1.5 on the X chromosome. (c) Were any double-crossover offspring expected? (d) Could the \(\mathrm{F}_{2}\) female offspring be used to construct the map? Why or why not?

Short answer

Short Answer: The genotypes of the F1 parents are: Mother \(X^+_wX^y_ct\) and Father \(X^Y_wX^+_c\). The constructed genetic map is: Yellow (\(0.894\)) - White (\(1.5\)) - Cut (\(2.639\)). Double-crossover offspring were expected in this cross. The F2 female offspring could not be used to construct the map because they would inherit a dominant allele from the F1 females, masking the expression of recessive alleles and preventing accurate observation of phenotypes.

Step-by-step solution

(a) Diagram the genotypes of the F1 parents

Since the F1 females are wild type for all three genes and no male carries a dominant allele, the genotypes of the F1 parents can be determined as follows: Mother (yellow-bodied, white-eyed, normal wings): \(X^+_wX^y_ct\) Father (normal-bodied, normal-eyed, cut wings): \(X^Y_wX^+_c\)

(b) Construct a genetic map

First, consider the number of recombinants and non-recombinants between each pair of genes: White-yellow: 112 recombinants + 949 non-recombinants = 1,061 total White-cut: 335 recombinants + 973 non-recombinants = 1,308 total Yellow-cut: 164 recombinants + 1,012 non-recombinants = 1,176 total Next, calculate the recombination frequencies: White-yellow: \(\frac{112}{1061} \times 100 = 10.56\%\) White-cut: \(\frac{335}{1308} \times 100 = 25.61\%\) Yellow-cut: \(\frac{164}{1176} \times 100 = 13.95\%\) Given that white is at locus 1.5 on the X chromosome, we can determine the locations of the other two genes as follows: Yellow: 1.5-(\({10.56\% \over 100}\)) = 0.894 Cut: 1.5+(\({13.95\% \over 100}\)) = 2.639 Thus, our genetic map is as follows: Yellow (\(0.894\)) - White (\(1.5\)) - Cut (\(2.639\))

(c) Were any double-crossover offspring expected?

Yes, double-crossover offspring were expected, as there were multiple crossing-overs between the three genes during meiosis. Double-crossover offspring can be identified as recombinants between the two outer genes, in this case, yellow and cut.

(d) Could the F2 female offspring be used to construct the map? Why or why not?

No, the F2 female offspring could not be used to construct the map, because they would inherit a dominant allele from the F1 females, and this dominant allele would mask the expression of the recessive alleles. As a result, we would not be able to observe the phenotypes of the F2 females and accurately establish the recombination frequencies between the genes.

Key concepts

Drosophila Genetics

Drosophila melanogaster, commonly known as the fruit fly, is one of the most valuable organisms in genetic research. It’s particularly favored due to its simple genetics, short life cycle, and the fact that it is easy to keep large populations.

Drosophila genetics involves understanding how traits are inherited in these flies. For instance, in an exercise involving Drosophila, you might see references to 'wild type', which implies the typical form of the fly as it occurs in nature, exhibiting the standard phenotype. Conversely, mutations can give rise to altered phenotypes, such as yellow-bodied or white-eyed flies.

When using Drosophila for genetic crosses, we observe the inheritance patterns of these traits to deduce information about the genes involved. In the textbook exercise, X-linked inheritance plays a significant role, given that the genes being observed are located on the X chromosome, which has implications for the resulting phenotypes of male and female offspring.

X-linked Genes

X-linked genes are those found on the X chromosome, one of the sex chromosomes that determine an organism's sex. In Drosophila, as with many other organisms, females have two X chromosomes (XX), and males have one X and one Y chromosome (XY).

This differential distribution of sex chromosomes means that males are hemizygous for X-linked genes; they have only one allele for each gene, inherited from their mother. Females, on the other hand, have two alleles for these genes, one from each parent.

The recessive, X-linked genes in the textbook exercise affect phenotypes like body color and eye color. Since males have only one X chromosome, any recessive allele on it will be expressed, even without a corresponding allele on the Y chromosome. Consequently, this explains why the F1 males in the exercise express the yellow-body and white-eye traits.

Recombination Frequency

Recombination frequency is a key measurement in genetics, quantifying the proportion of offspring that are recombinant, meaning their genotype reflects a new combination of alleles not seen in the parents. This recombination occurs due to crossovers between homologous chromosomes during meiosis, which is the process that produces gametes.

In genetic mapping, the recombination frequency is used to infer the physical distance between genes on a chromosome. The idea is based on the principle that the closer two genes are, the less likely it is for a crossover to occur between them, resulting in a lower recombination frequency. Conversely, genes situated further apart have a higher chance of recombination occurring between them.

The exercise provides recombination frequencies for different gene pairs, giving students the opportunity to calculate the distances between genes on the Drosophila X chromosome. These frequencies directly relate to the gene map they construct, and understanding this relationship is critical to solving the exercise.

Gene Mapping in Fruit Flies

Gene mapping in fruit flies involves determining the relative positions of genes on a chromosome. It is essential for understanding genetic linkage and recombination, as well as for studying various genetic phenomena.

In the context of the textbook exercise, we use recombination frequencies to construct a genetic map of three X-linked genes in Drosophila. This map provides us with a visual representation of gene locations and a way to predict offspring phenotypes based on known parental genotypes. Mapping genes can also reveal the existence of linkage groups, where genes are inherited together, and recombinational hotspots, regions where crossing-over is more likely to occur.

To create a genetic map, one starts by crossing individuals with different genotypes and observing the traits of the progeny. Then, by tallying recombinant and non-recombinant offspring, we calculate recombination frequencies. These values serve as an indirect measure of genetic distance, which is used to arrange the genes in a linear order. In the case of the exercise, the genetic map positions the yellow, white, and cut gene loci on the Drosophila X chromosome.