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Chapter 4: Problem 25

In Drosophila, an X-linked recessive mutation, scalloped (sd), causes irregular wing margins. Diagram the \(F_{1}\) and \(\mathrm{F}_{2}\) results if (a) a scalloped female is crossed with a normal male; (b) a scalloped male is crossed with a normal female. Compare these results with those that would be obtained if the scalloped gene were autosomal.

Short Answer

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Answer: In scenario (a), when a scalloped female is crossed with a normal male, all female offspring carry the scalloped allele (XX(sd)) while males express the scalloped phenotype (XY(sd)). In scenario (b), when a scalloped male is crossed with a normal female, half of the female offspring carry the scalloped allele (XX(sd+)) and the other half do not (XX(sd)). For both scenarios, the \(F_{2}\) generation shows different inheritance patterns compared to autosomal inheritance. If the scalloped gene were autosomal, the \(F_{1}\) generation would consist of 100% heterozygous offspring, while the \(F_{2}\) generation would result in a 3:1 phenotypic ratio (normal:scalloped), which is different from the observed inheritance pattern in X-linked inheritance.

Step by step solution

01

Understand X-linked inheritance

X-linked inheritance is a mode of inheritance for genes located on the X chromosome. In Drosophila, females have two X chromosomes (XX) and males have one X and one Y chromosome (XY). Recessive X-linked mutations will only be expressed in males, as they have only one X chromosome, but females need to have two copies of the recessive allele in order to express the trait.
02

Setup the parental genotypes

In scenario (a), the female is scalloped while the male is normal. If we represent the scalloped allele as "sd", this would mean: female = XX(sd) and male = XY(sd+). In scenario (b), the male is scalloped while the female is normal. This would mean: female = XX(sd+) and male = XY(sd).
03

Use Punnett squares to predict \(F_{1}\) generation

For both scenarios, we will use the Punnett square to predict the genotypes of the \(F_{1}\) generation. Scenario (a): | | X(sd+) | Y | |--------|-------|---------| | X(sd) | XX(sd)| XY(sd) | Scenario (b): | | X(sd+) | X(sd+) | |--------|-------|---------| | X(sd) | XX(sd)| XX(sd+) | In scenario (a), all female offspring would carry the scalloped allele (XX(sd)) while males would express the scalloped phenotype (XY(sd)). In scenario (b), half of the female offspring would carry the scalloped allele (XX(sd+)) and the other half would not (XX(sd)).
04

Predict \(F_{2}\) generation

We will now use the same approach to predict the \(F_{2}\) generation. Here, we will cross two heterozygous offspring from scenario (a) and two heterozygous offspring from scenario (b). Scenario (a): XX(sd) x XY(sd) = No change in \(F_{2}\), as all females will still carry the scalloped allele (XX(sd)) and all males will express the scalloped phenotype (XY(sd)). Scenario (b): XX(sd+) x XY(sd) = | | X(sd+) | X(sd+) | |--------|-------|---------| | X(sd) | XX(sd+)| XX(sd+) | | X(sd+) | XY(sd)| XY(sd+) | In the \(F_{2}\) generation, the ratio is 1:1 for normal and scalloped females and 1:1 for normal and scalloped males, as shown in the Punnett square.
05

Compare to autosomal inheritance

If the scalloped gene were autosomal, the inheritance pattern would be different. Assuming the female parent has the genotype sd/sd and the male parent has the genotype sd+/sd+, the \(F_{1}\) generation would consist of 100% heterozygous offspring (sd/sd+). The \(F_{2}\) generation from crossing two heterozygous offspring would result in a 1:2:1 genotypic ratio (sd/sd:sd/sd+:sd+/sd+) and a 3:1 phenotypic ratio (normal:scalloped). This is different from the observed inheritance pattern in X-linked inheritance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drosophila Genetics
Drosophila melanogaster, commonly known as the fruit fly, has been a cornerstone organism in genetic studies. Its genetic structure and significant number of mutant strains have made it an ideal model for studying inheritance patterns, including X-linked and autosomal inheritance. One important mutation studied in Drosophila is the scalloped (sd) mutation, which affects wing shape. Researchers often use Drosophila to illustrate how traits linked to sex chromosomes behave differently from those on autosomes鈥攃hromosomes that are not involved in determining sex. Studying these tiny flies helps scientists understand not only gene inheritance but also the broader concepts of genetics, evolution, and biological diversity.
  • Easy to breed and maintains a large number of offspring.
  • Possesses only four pairs of chromosomes.
  • Short life cycle, allowing for rapid observation of multiple generations.
Drosophila genetics also highlights critical processes like meiosis and genetic recombination, fundamental for advancing our understanding of heredity and variation.
Recessive Mutation
A recessive mutation refers to changes in a gene that are not expressed in the presence of a dominant allele. In the case of the scalloped (sd) mutation in Drosophila, this trait is recessive and X-linked. A mutation in such a gene requires that both alleles in females (XX) or the sole allele in males (XY) be mutated for the trait to be visible. Recessive traits often remain hidden from generation to generation, only to appear when the right alleles are inherited.

For X-linked recessive mutations, this generally means:
  • Females need two copies of the recessive allele to show the trait.
  • Males only need one copy because they only have one X chromosome.
In Drosophila, understanding how these mutations express themselves differently between sexes provides insight into genetic disorders in humans and other organisms as well, where X-linked recessive mutations can cause significant health issues.
Autosomal Inheritance
Autosomal inheritance refers to the transmission of genes located on autosomes, which are the chromosomes not involved in determining sex. Unlike X-linked inheritance, where the sex of the individual can influence the expression of a trait, autosomal inheritance usually does not differ between males and females. When considering a recessive mutation within an autosomal gene, both alleles in an individual must be mutated for the trait to be expressed.

Assuming a simple scenario with a recessive autosomal mutation:
  • Parental Generation: One parent homozygous recessive (e.g., sd/sd) and the other heterozygous or homozygous dominant (e.g., sd+/sd+).
  • Offspring would be heterozygous (sd/sd+) thus carriers of the trait but typically show the dominant phenotype.
  • When two carriers mate, potential outcomes for their offspring include a 1:2:1 genotypic ratio (homozygous recessive: heterozygous: homozygous dominant) and a 3:1 phenotypic ratio.
The difference in pattern seen here versus X-linked inheritance signifies the profound impact a chromosome's role and structure can have on heredity.

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Most popular questions from this chapter

Pigment in mouse fur is only produced when the \(C\) allele is present. Individuals of the \(c c\) genotype are white. If color is present, it may be determined by the \(A, a\) alleles. AA or \(A a\) results in agouti color, while aa results in black coats. (a) What \(F_{1}\) and \(F_{2}\) genotypic and phenotypic ratios are obtained from a cross between \(A A C C\) and aacc mice? (b) In three crosses between agouti females whose genotypes were unknown and males of the aacc genotype, the following phenotypic ratios were obtained:

In rats, the following genotypes of two independently assorting autosomal genes determine coat color: A third gene pair on a separate autosome determines whether or not any color will be produced. The \(C C\) and \(C c\) genotypes allow color according to the expression of the \(A\) and \(B\) alleles. However, the \(c c\) genotype results in albino rats regardless of the \(A\) and \(B\) alleles present. Determine the \(F_{1}\) phenotypic ratio of the following crosses: (a) \(A A b b C C \quad \times \quad\) aaBBcc (b) \(A a B B C C \quad \times \quad A A B b c c\) (c) \(A a B b C c \quad \times \quad\) AaBbcc (d) \(A a B B C c \quad \times \quad A a B B C c\) (e) \(A A B b C c \quad \times \quad A A B b c c\)

Discuss how temperature influences phenotypic expression.

In foxes, two alleles of a single gene, \(P\) and \(p,\) may result in lethality \((P P),\) platinum coat \((P p),\) or silver coat \((p p) .\) What ratio is obtained when platinum foxes are interbred? Is the \(P\) allele behaving dominantly or recessively in causing (a) lethality; (b) platinum coat color?

In Dexter and Kerry cattle, animals may be polled (hornless) or horned. The Dexter animals have short legs, whereas the Kerry animals have long legs. When many offspring were obtained from matings between polled Kerrys and horned Dexters, half were found to be polled Dexters and half polled Kerrys. When these two types of \(\mathrm{F}_{1}\) cattle were mated to one another, the following \(\mathrm{F}_{2}\) data were obtained: \(3 / 8\) polled Dexters \(3 / 8\) polled Kerrys \(1 / 8\) horned Dexters \(1 / 8\) horned Kerrys A geneticist was puzzled by these data and interviewed farmers who had bred these cattle for decades. She learned thatKerrys were true breeding. Dexters, on the other hand, were not true breeding and never produced as many offspring as Kerrys. Provide a genetic explanation for these observations.
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