In Figure 6-1,

In Figure 6-2, explain how the mutant polypeptide acts as a spoiler and what its net effect on phenotype is.

In Figure 6-6, assess the allele V^f with respect to the V^by allele: is it dominant? recessive? codominant? incompletely dominant?

In Figure 6-11,

In Figure 6-12,

In Figure 6-13, explain at the protein level why this heterokaryon can grow on minimal medium.

In Figure 6-14, write possible genotypes for each of the four snakes illustrated.

In Figure 6-15,

In Figure 6-16, if you selfed 10 different F₂ pink plants, would you expect to find any white-flowered plants among the offspring? Any blue-flowered plants?

In Figure 6-19,

In Figure 6-21, propose a specific genetic explanation for individual Q (give a possible genotype, defining the alleles).

In humans, the disease galactosemia causes mental retardation at an early age. Lactose (milk sugar) is broken down to galactose plus glucose. Normally, galactose is broken down further by the enzyme galactose-1-phos-phate uridyltransferase (GALT). However, in patients with galactosemia, GALT is inactive, leading to a buildup of high levels of galactose, which, in the brain, causes mental retardation. How would you provide a secondary cure for galactosemia? Would you expect this disease phenotype to be dominant or recessive?

In humans, PKU (phenylketonuria) is a recessive disease caused by an enzyme inefficiency at step A in the following simplified reaction sequence, and AKU (alkaptonuria) is another recessive disease due to an enzyme inefficiency in one of the steps summarized as step B here:

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A person with PKU marries a person with AKU. What phenotypes do you expect for their children? All normal, all having PKU only, all having AKU only, all having both PKU and AKU, or some having AKU and some having PKU?

In Drosophila, the autosomal recessive bw causes a dark brown eye, and the unlinked autosomal recessive st causes a bright scarlet eye. A homozygote for both genes has a white eye. Thus, we have the following correspondences between genotypes and phenotypes:

Construct a hypothetical biosynthetic pathway showing how the gene products interact and why the different mutant combinations have different phenotypes.

Several mutants are isolated, all of which require compound G for growth. The compounds (A to E) in the biosynthetic pathway to G are known, but their order in the pathway is not known. Each compound is tested for its ability to support the growth of each mutant (1 to 5). In the following table, a plus sign indicates growth and a minus sign indicates no growth.

In a certain plant, the flower petals are normally purple. Two recessive mutations arise in separate plants and are found to be on different chromosomes. Mutation 1 (m₁) gives blue petals when homozygous (m₁/m₁). Mutation 2 (m₂) gives red petals when homozygous (m₂/m₂). Biochemists working on the synthesis of flower pigments in this species have already described the following pathway:

In sweet peas, the synthesis of purple anthocyanin pigment in the petals is controlled by two genes, B and D. The pathway is

If a man of blood-group AB marries a woman of blood-group A whose father was of blood-group O, to what different blood groups can this man and woman expect their children to belong?

Most of the feathers of erminette fowl are light colored, 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 hypotheses?

Radishes may be long, round, or oval, and they may be red, white, or purple. You cross a long, white variety with a round, red one and obtain an oval, purple F₁. The F₂ shows nine phenotypic classes as follows: 9 long, red; 15 long, purple; 19 oval, red; 32 oval, purple; 8 long, white; 16 round, purple; 8 round, white; 16 oval, white; and 9 round, red.

In the multiple-allele series that determines coat color in rabbits, c⁺ encodes agouti, c^ch encodes chinchilla (a beige coat color), and c ^h encodes 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?

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Black, sepia, cream, and albino are coat colors of guinea pigs. Individual animals (not necessarily from pure lines) showing these colors were intercrossed; the results are tabulated as follows, where the abbreviations A (albino), B (black), C (cream), and S (sepia) represent 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.

Consider two blood polymorphisms that humans have in addition to the ABO system. Two alleles L^M and L^N determine the M, N, and MN blood groups. The dominant allele R of a different gene causes a person to have the Rh+ (rhesus positive) phenotype, whereas the homozygote for r is Rh^— (rhesus negative). Two men took a paternity dispute to court, each claiming three children to be his own. The blood groups of the men, the children, and their mother were as follows:

From this evidence, can the paternity of the children be established?

On a fox ranch in Wisconsin, a mutation arose that gave a “platinum” coat color. The platinum color proved very popular with buyers of fox coats, but the breeders could not develop a pure-breeding platinum strain. Every time two platinums were crossed, some normal foxes appeared in the progeny. For example, the repeated matings of the same pair of platinums produced 82 platinum and 38 normal progeny. All other such matings gave similar progeny ratios. State a concise genetic hypothesis that accounts for these results.

For several years, Hans Nachtsheim investigated an inherited anomaly of the white blood cells of rabbits. This anomaly, termed the Pelger anomaly, is the arrest of the segmentation of the nuclei of certain white cells. This anomaly does not appear to seriously burden the rabbits.

Two normal-looking fruit flies were crossed, and, in the progeny, there were 202 females and 98 males.

You have been given a virgin Drosophila female. You notice that the bristles on her thorax are much shorter than normal. You mate her with a normal male (with long bristles) and obtain the following F₁ progeny: short-bristled females, long-bristled females, and long-bristled males. A cross of the F₁ long-bristled females with their brothers gives only long-bristled F₂. A cross of short-bristled females with their brothers gives short-bristled females, long-bristled females, and long-bristled males. Provide a genetic hypothesis to account for all these results, showing genotypes in every cross.

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A dominant allele H reduces the number of body bristles that Drosophila flies have, giving rise to a “hairless” phenotype. In the homozygous condition, H is lethal. An independently assorting dominant allele S has no effect on bristle number except in the presence of H, in which case a single dose of S suppresses the hairless phenotype, thus restoring the hairy phenotype. However, S also is lethal in the homozygous (S/S) condition.

After irradiating wild-type cells of Neurospora (a haploid fungus), a geneticist finds two leucine-requiring auxotrophic mutants. He combines the two mutants in a heterokaryon and discovers that the heterokaryon is prototrophic.

A yeast geneticist irradiates haploid cells of a strain that is an adenine-requiring auxotrophic mutant, caused by mutation of the gene adel. Millions of the irradiated cells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies. These colonies are crossed individually with a wild-type strain. Two types of results are obtained:

(1) prototroph × wild type : progeny all prototrophic

(2) prototroph × wild type : progeny 75% prototrophic, 25% adenine-requiring auxotrophs

In roses, the synthesis of red pigment is by two steps in a pathway, as follows:

Because snapdragons (Antirrhinum) possess the pigment anthocyanin, they have reddish purple petals. Two pure anthocyaninless 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.

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 fowl 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.

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 pet als, and the second line had white petals. The following crosses were made between pure lines, with the results shown:

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A woman who owned a purebred albino poodle (an autosomal recessive phenotype) wanted white puppies; so 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, all of them were 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?

A snapdragon plant that bred true for white petals was crossed with a plant that bred true for purple petals, and all the F₁ had white petals. The F₁ was selfed. Among the F₂, three phenotypes were observed in the following numbers:

What were the genotypes of the F₂ plants crossed?

Most flour beetles are black, but several color variants are known. Crosses of pure-breeding parents produced the following results (see table) 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.

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Two albinos marry and have four normal children. How is this possible?

Consider the production of flower color in the Japanese morning glory (Pharbitis nil). Dominant alleles of either of two separate genes (A/− ·b/b or a/a · B/−) produce purple petals. A/− · B/− produces blue petals, and a/a · b/b produces scarlet petals. Deduce the genotypes of parents and progeny in the following crosses:

Corn breeders obtained pure lines whose kernels turn sun red, pink, scarlet, or orange when exposed to sunlight (normal kernels remain yellow in sunlight). Some crosses between these lines produced the following results. The phenotypes are abbreviated O, orange; P, pink; Sc, scarlet; and SR, sun red.

Analyze the results of each cross, and provide a unifying hypothesis to account for all the results. (Explain all symbols that you use.)

Many kinds of wild animals have the agouti coloring pattern, in which each hair has a yellow band around it.

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 yellow, gray, black, and white. Determine the genotypes of the parents.

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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 illustrations). Assume independent assortment.

The production of eye-color pigment in Drosophila requires the dominant allele A. The dominant allele P of a second independent gene turns the pigment to purple, but its recessive allele leaves it red. A fly producing no pigment has white eyes. Two pure lines were crossed with the following results:

Explain this mode of inheritance, and show the genotypes of the parents, the F₁, and the F₂.

When true-breeding brown dogs are mated with certain true-breeding white dogs, all the F₁ pups are white. The F₂ progeny from some F₁ × F₁ crosses were 118 white, 32 black, and 10 brown pups. What is the genetic basis for these results?

Wild-type strains of the haploid fungus Neurospora can make their own tryptophan. An abnormal allele td renders the fungus incapable of making its own tryptophan. An individual of genotype td grows only when its medium supplies tryptophan. The allele su assorts independently of td; its only known effect is to suppress the td phenotype. Therefore, strains carrying both td and su do not require tryptophan for growth.

Mice of the genotypes A/A; B/B; C/C; D/D; S/S and a/a; b/b; c/c; d/d; s/s are crossed. The progeny are intercrossed. What phenotypes will be produced in the F₂ and in what proportions? [The allele symbols stand for the following: A = agouti, a = solid (nonagouti); B = black pigment, b = brown; C = pigmented, c = albino; D = nondilution, d = dilution (milky color); S = unspotted, s = pigmented spots on white background.]

Consider the genotypes of two lines of chickens: the pure-line mottled Honduran is i/i; D/D; M/M; W/W, and the pure-line leghorn is I/I; d/d; m/m; w/w, where

These four genes assort independently. Starting with these two pure lines, what is the fastest and most convenient way of generating a pure line that has colored feathers, has a simplex comb, is beardless, and has yellow skin? Make sure that you show

The following pedigree is for a dominant phenotype governed by an autosomal allele. What does this pedigree suggest about the phenotype, and what can you deduce about the genotype of individual A?

Petal coloration in foxgloves is determined by three genes. M encodes an enzyme that synthesizes anthocyanin, the purple pigment seen in these petals; m/m produces no pigment, resulting in the phenotype albino with yellowish spots. D is an enhancer of anthocyanin, resulting in a darker pigment; d/d does not enhance. At the third locus, w/w allows pigment deposition in petals, but W prevents pigment deposition except in the spots and so results in the white, spotted phenotype. Consider the following two crosses:

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In each case, give the genotypes of parents and progeny with respect to the three genes.

In one species of Drosophila, the wings are normally round in shape, but you have obtained two pure lines, one of which has oval wings and the other sickle-shaped wings. Crosses between pure lines reveal the following results:

Mice normally have one yellow band on each hair, but variants with two or three bands are known. A female mouse having one band was crossed with a male having three bands. (Neither animal was from a pure line.) The progeny were

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:

In Drosophila, an autosomal gene determines the shape of the hair, with B giving straight and b giving bent hairs. On another autosome, there is a gene of which a dominant allele I inhibits hair formation so that the fly is hairless (i has no known phenotypic effect).

The following pedigree concerns eye phenotypes in Tribolium beetles. The solid symbols represent black eyes, the open symbols represent brown eyes, and the cross symbols (X) represent the “eyeless” phenotype, in which eyes are totally absent.

A plant believed to be heterozygous for a pair of alleles B/b (where B encodes yellow and b encodes bronze) was selfed, and, in the progeny, there were 280 yellow and 120 bronze plants. Do these results support the hypothesis that the plant is B/b?

A plant thought to be heterozygous for two independently assorting genes (P/p; Q/q) was selfed, and the progeny were

Do these results support the hypothesis that the original plant was P/p; Q/q?

A plant of phenotype 1 was selfed, and, in the progeny, there were 100 plants of phenotype 1 and 60 plants of an alternative phenotype 2. Are these numbers compatible with expected ratios of 9:7, 13:3, and 3:1? Formulate a genetic hypothesis on the basis of your calculations.

Four homozygous recessive mutant lines of Drosophüa melanogaster (labeled 1 through 4) showed abnormal leg coordination, which made their walking highly erratic. These lines were intercrossed; the phenotypes of the F₁ flies are shown in the following grid, in which “+” represents wild-type walking and “−‣ represents 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 figure below).

A pure-breeding strain of squash that produced diskshaped fruits (see the accompanying illustration) 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.

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 from a ruptured aorta.)

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

In corn, three dominant alleles, called A, C, and R, must be present to produce colored seeds. Genotype A/−; C/−; R/− is colored; all others are colorless. A colored plant is crossed with three tester plants of known genotype. With tester a/a; c/c; R/R, the colored plant produces 50 percent colored seeds; with a/a; C/C; r/r, it produces 25 percent colored; and with A/A; c/c; r/r, it produces 50 percent colored. What is the genotype of the colored plant?

The production of pigment in the outer layer of seeds of corn requires each of the three independently assorting genes A, C, and R to be represented by at least one dominant allele, as specified in Problem 64. The dominant allele Pr of a fourth independently assorting gene is required to convert the biochemical precursor into a purple pigment, and its recessive allele pr makes the pigment red. Plants that do not produce pigment have yellow seeds. Consider a cross of a strain of genotype A/A; C/C; R/R; pr/pr with a strain of genotype a/a; c/c; r/r; Pr/Pr.

The allele B gives mice a black coat, and b gives a brown one. The genotype e/e of another, independently assorting gene prevents the expression of B and b, making the coat color beige, whereas E/- permits the 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.

A researcher crosses two white-flowered lines of Antirrhinum plants as follows and obtains the following results:

Assume that two pigments, red and blue, mix to give the normal purple color of petunia petals. Separate bio chemical 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 is normally at a concentration too low to color petals.

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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 into the yellow intermediate of the red pathway.

In the diagram, the letters A through 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 encode enzyme function and that recessive alleles result in a lack of enzyme function. Deduce which combinations of true-breeding parental genotypes could be crossed to produce F₂ progeny in the following ratios:

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

The flowers of nasturtiums (Tropaeolum majus) may be single (S), double (D), or superdouble (Sd). Superdoubles are female sterile; they originated from a double-flowered variety. Crosses between varieties gave the progeny listed in the following table, in which pure means “pure breeding.”

Using your own genetic symbols, propose an explanation for these results, showing

In a certain species of fly, the normal eye color is red (R). Four abnormal phenotypes for eye color were found: two were yellow (Y1 and Y2), one was brown (B), and one was orange (O). A pure line was established for each phenotype, and all possible combinations of the pure lines were crossed. Flies of each F₁ were intercrossed to produce an F₂. The F₁ and the F₂ flies are shown within the following square; the pure lines are given at the top and at the left-hand side.

In common wheat, Triticum aestivum, kernel color is determined by multiply duplicated genes, each with an R and an r allele. Any number of R alleles will give red, and a complete lack of R alleles will give the white phenotype. In one cross between a red pure line and a white pure line, the F₂ was red and white.

The following pedigree shows the inheritance of deafmutism.

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The pedigree below is for blue sclera (bluish thin outer wall of the eye) and brittle bones.

Workers of the honeybee line known as Brown (nothing to do with color) show what is called “hygienic behavior”; that is, they uncap hive compartments containing dead pupae and then remove the dead pupae. This behavior prevents the spread of infectious bacteria through the colony. Workers of the Van Scoy line, however, do not perform these actions, and therefore this line is said to be “nonhygienic.” When a queen from the Brown line was mated with Van Scoy drones, all the F₁ were nonhygienic. When drones from this F₁ inseminated a queen from the Brown line, the progeny behaviors were as follows:

However, when the compartment of dead pupae was uncapped by the beekeeper and the nonhygienic honeybees were examined further, about half the bees were found to remove the dead pupae, but the other half did not.

(Note: Workers are sterile, and all bees from one line carry the same alleles.)

The normal color of snapdragons is red. Some pure lines showing variations of flower color have been found. When these pure lines were crossed, they gave the following results (see the table):

Consider the following F₁ individuals in different species and the F₂ ratios produced by selfing:

If each F₁ were testcrossed, what phenotypic ratios would result in the progeny of the testcross?

To understand the genetic basis of locomotion in the diploid nematode Caenorhabditis elegans, recessive mutations were obtained, all making the worm “wiggle” ineffectually instead of moving with its usual smooth gliding motion. These mutations presumably affect the nervous or muscle systems. Twelve homozygous mutants were intercrossed, and the F₁ hybrids were examined to see if they wiggled. The results were as follows, where a plus sign means that the F₁ hybrid was wild type (gliding) and “w” means that the hybrid wiggled.

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A geneticist working on a haploid fungus makes a cross between two slow-growing mutants called mossy and spider (referring to the abnormal appearance of the colonies). Tetrads from the cross are of three types (A, B, C), but two of them contain spores that do not germinate.

Devise a model to explain these genetic results, and propose a molecular basis for your model.

In the nematode C. elegans, some worms have blistered cuticles due to a recessive mutation in one of the bli genes. Someone studying a suppressor mutation that suppressed bli-3 mutations wanted to know if it would also suppress mutations in bli-4. They had a strain that was homozygous for this recessive suppressor mutation, and its phenotype was wild type.