Ch. 4 - Modification of Mendelian Ratios

Klug - Essentials of Genetics 10th Edition

Klug10th EditionEssentials of GeneticsISBN: 9780135588789Not the one you use?Change textbook

Klug 10th Edition

Ch. 4 - Modification of Mendelian Ratios

Problem 23

Chapter 4, Problem 23

Three autosomal recessive mutations in Drosophila, all with tan eye color (r1, r2, and r3), are independently isolated and subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—

Verified step by step guidance

Step 1: Understand the concept of complementation analysis. When two mutations are crossed, if the offspring show the wild-type phenotype, the mutations complement each other and are likely in different genes. If the offspring show the mutant phenotype, the mutations fail to complement and are likely alleles of the same gene.

Step 2: Identify the crosses given in the problem and note the phenotypes of the offspring for each cross between the mutations r1, r2, and r3. Determine which crosses show complementation (wild-type eyes) and which show failure to complement (tan eyes).

Step 3: Use the complementation results to group the mutations. Mutations that fail to complement each other belong to the same complementation group (i.e., are alleles of the same gene). Mutations that complement each other belong to different genes.

Step 4: For the cross that is not shown, predict the phenotype of the offspring based on the complementation groups established. If the two mutations are in the same group, expect mutant phenotype; if in different groups, expect wild-type phenotype.

Step 5: Summarize which mutations are alleles of one another based on the complementation groups and state the predicted phenotype for the missing cross accordingly.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Autosomal Recessive Mutations

Autosomal recessive mutations require two copies of the mutant allele for the phenotype to be expressed. In Drosophila, these mutations are located on non-sex chromosomes and typically result in a loss-of-function phenotype, such as tan eye color in this case.

Complementation Analysis

Complementation analysis tests whether two mutations causing similar phenotypes are in the same gene (allelic) or in different genes. If two mutations complement, the heterozygote shows a wild-type phenotype, indicating mutations are in different genes; if not, they are alleles of the same gene.

Allelic Relationships and Predicting Cross Outcomes

Alleles are different versions of the same gene. When crossing two mutants, if they fail to complement, the offspring show the mutant phenotype, indicating allelism. Predicting results of untested crosses involves applying complementation principles to infer whether mutations are allelic or not.

Recommended video:

Guided course

26:08

Trihybrid Cross

Related Practice

Textbook Question

Consider the three pedigrees below, all involving a single human trait.

For each combination that you excluded, indicate the single individual in generation II (e.g., II-1, II-2) that was most instrumental in your decision to exclude it. If none were excluded, answer 'none apply.'

572

views

Textbook Question

Consider the three pedigrees below, all involving a single human trait.

Given your conclusions in part (a), indicate the genotype of the following individuals: II-1, II-6, II-9 If more than one possibility applies, list all possibilities. Use the symbols A and a for the genotypes.

1023

views

Textbook Question

Labrador retrievers may be black, brown (chocolate), or golden (yellow) in color (see chapter-opening photo). While each color may breed true, many different outcomes are seen when numerous litters are examined from a variety of matings where the parents are not necessarily true breeding. Following are just some of the many possibilities.

(a) black x brown → all black

(b) black x brown → 1/2 black, 1/2 brown

(d) black x golden → all black

(e) black x golden → 4/8 golden 3/8 black 1/8 brown

(f) black x golden → 2/4 golden 1/4 black 1/4 brown

(g) brown x brown → 3/4 brown 1/4 golden

(h) black x black → 9/16 black 4/16 golden, 3/16 brown

Propose a mode of inheritance that is consistent with these data, and indicate the corresponding genotypes of the parents in each mating. Indicate as well the genotypes of dogs that breed true for each color.

1012

views

Textbook Question

Horses can be cremello (a light cream color), chestnut (a brownish color), or palomino (a golden color with white in the horse's tail and mane). Of these phenotypes, only palominos never breed true.

From the results given above, determine the mode of inheritance by assigning gene symbols and indicating which genotypes yield which phenotypes.

1086

views

Textbook Question

Horses can be cremello (a light cream color), chestnut (a brownish color), or palomino (a golden color with white in the horse's tail and mane). Of these phenotypes, only palominos never breed true.

Predict the F₁ and F₂ results of many initial matings between cremello and chestnut horses.

546

views

Textbook Question

Pigment in mouse fur is only produced when the C allele is present. Individuals of the cc genotype are white. If color is present, it may be determined by the A, a alleles. AA or Aa results in agouti color, while aa results in black coats. What F₁ and F₂ genotypic and phenotypic ratios are obtained from a cross between AACC and aacc mice?

751

views