Both recombination and complementation can take two different recessive mutants and from them produce a wild type. How would you explain the difference between these two processes?

Erminette fowls have mostly light-colored feathers with an occasional black one, giving a flecked appearance. A cross of two erminettes produced a total of 48 progeny, consisting of 22 erminettes, 14 blacks, and 12 pure whites. What genetic basis of the erminette pattern is suggested? How would you test your hypothesis?

In the multiple allelic series that determines coat color in rabbits, c⁺ codes for agouti, c^ch is chinchilla, and c^h is Himalayan. Dominance is in the order c⁺ > c^ch > c^h. In a cross of c⁺/c^ch×c^ch/c^h, what proportion of progeny will be chinchilla?

Black, sepia, cream, and albino are all coat colors of laboratory guinea pigs. Individual animals (not necessarily from pure lines) showing these colors were intercrossed; the results are tabulated below, where we are using the abbreviation A for albino; B, black; C, cream; and S, sepia, for the phenotypes:

In a maternity ward, four babies become accidentally mixed up. The ABO types of the four babies are known to be O, A, B, and AB. The ABO types of the four sets of parents are determined. Indicate which baby belongs to each set of parents: (a) AB×O, (b) A×O, (c) A×AB, (d) O×O.

Two fruit flies of wild-type phenotype were crossed, and in the progeny there were 202 females and 98 males.
a. What is unusual about this result?

b. Provide a genetic explanation.

c. Provide a test of your hypothesis.

A pure-breeding strain of squash that produced disk-shaped fruits (see the figure below) was crossed with a pure-breeding strain having long fruits. The F₁ had disk fruits, but the F₂ showed a new phenotype, sphere, and was composed of the following proportions:

Propose an explanation for these results, and show the genotypes of the P, F₁, and F₂ generations.

(Illustration from P. J. Russell, Genetics, 3d ed. Harper- Collins, 1992.)

Wild-type snapdragons (Antirrhinum) make a red anthocyanin pigment and have red petals. Two pure anthocyanin-less mutant lines of Antirrhinum were developed, one in California and one in Holland. They looked identical in having no red pigment at all, manifested as white (albino) flowers. However, when petals from the two lines were ground up together in buffer in the same test tube, the solution, which appeared colorless at first, gradually turned red.
a. What control experiments should an investigator conduct before proceeding with further analysis?

b. What could account for the production of the red color in the test tube?

c. According to your explanation for part b, what would be the genotypes of the two lines?

d. If the two white lines are crossed, what would you predict the phenotypes of the F₁ and F₂ to be? What genetic principle is illustrated by the F₁ results? The F₂ results?

Four homozygous recessive mutant lines of Drosophila melanogaster (labeled 1 through 4) showed abnormal leg coordination, which made walking highly erratic. These lines were intercrossed; the phenotypes of the F₁ animals are shown in the following grid in which "+" means the F₁ showed wild-type walking and "" means abnormal walking:

Three independently isolated tryptophan-requiring mutants of haploid yeast are called trpB, trpD, and trpE. Cell suspensions of each are streaked on a plate of nutritional medium supplemented with just enough tryptophan to permit weak growth for a trp strain. The streaks are arranged in a triangular pattern so that they do not touch one another. Luxuriant growth is noted at both ends of the trpE streak and at one end of the trpD streak (see the following figure).

The frizzle fowl is much admired by poultry fanciers. It gets its name from the unusual way that its feathers curl up, giving the impression that it has been (in the memorable words of animal geneticist F. B. Hutt) "pulled backwards through a knothole." Unfortunately, frizzle fowls do not breed true; when two frizzles are intercrossed, they always produce 50 percent frizzles, 25 percent normal, and 25 percent with peculiar woolly feathers that soon fall out, leaving the birds naked.
a. Give a genetic explanation for these results, showing genotypes of all phenotypes, and provide a statement of how your explanation works.

b. If you wanted to mass-produce frizzle fowls for sale, which types would be best to use as a breeding pair?

Marfan's syndrome is a disorder of the fibrous connective tissue characterized by many symptoms, including long, thin digits, eye defects, heart disease, and long limbs. (Flo Hyman, the American volleyball star, suffered from Marfan's syndrome. She died soon after a game from a ruptured aorta.)
a. Use the pedigree shown above to propose a mode of inheritance for Marfan's syndrome.

b. What genetic phenomenon is shown by this pedigree?

c. Speculate on a reason for such a phenomenon.

(Illustration from J. V. Neel and W. J. Schull, Human Heredity. University of Chicago Press, 1954.)

A woman owned a fine purebred albino female poodle (an autosomal recessive phenotype) and wanted to mate her to produce a litter of white puppies. She took the dog to a breeder, who said he would mate the female with an albino stud male, also from a pure stock. When six puppies were born, they were all black, so the woman sued the breeder, claiming that he replaced the stud male with a black dog, giving her six unwanted puppies. You are called in as an expert witness, and the defense asks you if it is possible to produce black offspring from two pure-breeding recessive albino parents. What testimony do you give?

The petals of the plant Collinsia parviflora are normally blue, giving the species its common name "blue-eyed Mary." Two pure-breeding lines were obtained from color variants found in nature; the first line had pink petals and the second line had white petals. The following crosses were made between pure lines, with the results shown:

Unpacking the Problem

a. What is the character being studied?

b. What is the wild-type phenotype?

c. What is a variant?

d. What are the variants in this problem?

e. What does "in nature" mean?

f. In what way would the variants have been found in nature? (Describe the scene.)

g. At which stages in the experiments would seeds be used?

h. Would the way of writing a cross "blue×white" (for example) mean the same as "white×blue"? Would you expect similar results? Why or why not?

i. In what way do the first two rows in the table differ from the third row?

j. Which phenotypes are dominant?

k. What is complementation?

l. Where does the blueness come from in the progeny of the pink×white cross?

m. What genetic phenomenon does the production of a blue F₁ from pink and white parents represent?

n. List any ratios you can see.

o. Are there any monohybrid ratios?

p. Are there any dihybrid ratios?

q. What does observing monohybrid and dihybrid ratios tell you?

r. List four modified Mendelian ratios that you can think of.

s. Are there any modified Mendelian ratios in the problem?

t. What do modified Mendelian ratios indicate generally?

u. What do the specific modified ratio or ratios in this problem indicate?

v. Draw chromosomes representing the meioses in the parents in the cross blue×white, and meiosis in the F₁.

w. Repeat for the cross blue×pink.

Most flour beetles are black, but several color variants are known. Crosses of pure-breeding parents produced the following results in the F₁ generation, and intercrossing the F₁ from each cross gave the ratios shown for the F₂ generation. The phenotypes are abbreviated Bl, black; Br, brown; Y, yellow; and W, white.

An allele A that is not lethal when homozygous causes rats to have yellow coats. The allele R of a separate gene that assorts independently produces a black coat. Together, A and R produce a grayish coat, whereas a and r produce a white coat. A gray male is crossed with a yellow female, and the F₁ is 3/8 yellow, 3/8 gray, 1/8 black, and 1/8 white. Determine the genotypes of the parents.

The genotype r/r;p/p gives fowl a single comb, R/;P/ gives a walnut comb, r/r;P/ gives a pea comb, and R/;p/p gives a rose comb (see the figure on the top of the facing page). Note that the genes are unlinked.

The allele B gives mice a black coat, and b gives a brown one. The genotype e/e for another, independently assorting gene prevents expression of B and b, making the coat color beige, whereas E/ permits expression of B and b. Both genes are autosomal. In the following pedigree, black symbols indicate a black coat, pink symbols indicate brown, and white symbols indicate beige:

Assume that two pigments, red and blue, mix to give the normal wild-type purple color of petunia petals. Separate biochemical pathways synthesize the two pigments, as shown in the top two rows of the accompanying diagram. "White" refers to compounds that are not pigments. (Total lack of pigment results in a white petal.) Red pigment forms from a yellow intermediate that normally is at a concentration too low to color petals. A third pathway whose compounds do not contribute pigment to petals normally does not affect the blue and red pathways, but if one of its intermediates (white₃) should build up in concentration, it can be converted to the yellow intermediate of the red pathway. In the diagram, A to E represent enzymes; their corresponding genes, all of which are unlinked, may be symbolized by the same letters.

Assume that wild-type alleles are dominant and code for enzyme function, and that recessive alleles represent lack of enzyme function. Deduce which combinations of true-breeding parental genotypes could be crossed to produce F₂ progenies in the following ratios:
a. 9 purple:3 green:4 blue

b. 9 purple:3 red:3 blue:1 white

c. 13 purple:3 blue

d. 9 purple:3 red:3 green:1 yellow

e. Which of the mutations is acting as a suppressor?

f. Which of the mutations is showing recessive epistasis?

(Note: blue mixed with yellow makes green; assume that no mutations are lethal.)

The following pedigree shows the inheritance of deaf-mutism:

The following pedigree is for blue sclera (bluish thin outer wall to the eye) and brittle bones:

In minks, wild types have an almost black coat. Breeders have developed many pure lines of color variants for the mink coat industry. Two such pure lines are platinum (blue-gray) and aleutian (steel-gray). These lines were used in crosses, with the following results:

A geneticist is interested in the genes that interact to synthesize leucine (an amino acid) in the haploid filamentous fungus Aspergillus. He treats haploid cells with ultraviolet light to increase the mutation rate and obtains five haploid leucinerequiring mutants (a to e), all of which need leucine to be added to the medium in order for them to grow. (Without leucine, none of them will grow.)

Experiment I. He first makes heterokaryons between them to check on their functional relationships. He obtains the following results, where "+" indicates that the heterokaryon grew, and "" that the heterokaryon did not grow, on a medium lacking leucine:

Experiment II. The geneticist then intercrosses the mutants in all possible combinations. From each cross, he tests 500 ascospore progeny by inoculating them onto a medium lacking leucine. The results are in the following table. The numbers represent the number of progeny out of 500 that could synthesize their own leucine (and grow on a medium containing no leucine):

Wild-type Neurospora (a haploid fungus) can synthesize its own adenine from inorganic components in the growth medium. In a mutational analysis of the synthetic pathway for making adenine, 10 adenine-requiring mutations of Neurospora were obtained, and heterokaryons were made in all pairwise combinations. The results are shown in the following table in which a plus sign means the heterokaryon grew and a negative sign means the heterokaryon did not grow:

Mutants 1,3, and 4 were tested for growth on the compounds CAIR, AIR, and SAICAR, all chemically related to adenine. The results were as follows, where "+" means growth and "," none:

Mutants 1, 3, and 4 were intercrossed, and 1000 ascospores were plated from each cross onto medium lacking adenine, to select for wild types. The results were as follows:

In a plant the gene e⁺ codes for a certain enzyme. The gene r⁺ codes for a regulatory protein that is needed to bind to the regulatory region of gene e⁺ in order for transcription to occur. A recessive null mutation is obtained in each of the genes, r and e, respectively. A dihybrid e⁺/e;r⁺/r is allowed to self. In the progeny what will be the proportion of individuals with enzyme? (Assume lack of enzyme is not lethal and that the genes are on different chromosomes.)

It is known that a certain Drosophila protein (call it P) is necessary for normal flight. This protein is known to be composed of only one polypeptide chain. Two pure lines of mutants (lines 1 and 2) were obtained that could not fly. When crossed to wild type, they both gave the same result:

When the two pure lines were intercrossed, the results were

Protein from all members of the F₂ was run on electrophoretic gels and stained using an antibody that is specific for protein P. It was found that the F₂ individuals could be grouped into six different types as follows:

(The thin bands represent very weak antibody staining.)
a. Provide a general explanation for the molecular results revealed by the gel analysis. In particular, state the difference in the molecular nature of the two original mutations.

b. According to your explanation, what are the genotypes of the six F₂ types found?

c. What is the expected frequency of type 4?

d. Which of the six F₂ gel types represent the 9/16 that are phenotypically wild-type?

e. Draw the expected gel pattern of the two parental mutants and the F₁.

Using the diagram, deduce genotype and dominance and predict the genotypes in the F₂s.

The wing covers of grain beetles exhibit one of three phenotypes, showing a diamond, a spot, or a stripe. The following crosses were made between various strains that were not necessarily pure-breeding: