In Figure 4-3, would there be any meiotic products that did not undergo a crossover in the meiosis illustrated? If so, what colors would they be in the color convention used?

In Figure 4-6, why does the diagram not show meioses in which two crossovers occur between the same two chromatids (such as the two inner ones)?

In Figure 4-8, some meiotic products are labeled parental. Which parent is being referred to in this terminology?

In Figure 4-9, why is only locus A shown in a constant position?

In Figure 4-10, what is the mean frequency of crossovers per meiosis in the region A-B? The region B–C?

In Figure 4-11, is it true to say that from such a cross the product v cv+ can have two different origins?

In Figure 4-14, in the bottom row four colors are labeled SCO. Why are they not all the same size (frequency)?

Using the conventions of Figure 4-15, draw parents and progeny classes from a cross

P M′″/p M′ × p M′/p M″″

In Figure 4-17, draw the arrangements of alleles in an octad from a similar meiosis in which the upper product of the first division segregated in an upside-down manner at the second division.

In Figure 4-19, what would be the RF between A/a and B/b in a cross in which purely by chance all meioses had four-strand double crossovers in that region?

A plant of genotype

is testcrossed with

If the two loci are 10 m.u. apart, what proportion of progeny will be AB/ab?

The A locus and the D locus are so tightly linked that no recombination is ever observed between them. If Ad/Ad is crossed with aD/aD and the F₁ is intercrossed, what phenotypes will be seen in the F₂ and in what proportions?

The R and S loci are 35 m.u. apart. If a plant of genotype

is selfed, what progeny phenotypes will be seen and in what proportions?

The cross E/E · F/F × e/e · f/f is made, and the F₁ is then backcrossed with the recessive parent. The progeny genotypes are inferred from the phenotypes. The progeny genotypes, written as the gametic contributions of the heterozygous parent, are in the following proportions:

Explain these results.

A strain of Neurospora with the genotype H · I is crossed with a strain with the genotype h · i. Half the progeny are H · I, and the other half are h · i. Explain how this outcome is possible.

A female animal with genotype A/a · B/b is crossed with a double-recessive male (a/a · b/b). Their progeny include 442 A/a · B/b, 458 a/a · b/b, 46 A/a · b/b, and 54 a/a · B/b. Explain these results.

If A/A · B/B is crossed with a/a · b/b and the F₁ is testcrossed, what percentage of the testcross progeny will be a/a · b/b if the two genes are (a) unlinked; (b) completely linked (no crossing over at all); (c) 10 m.u. apart; (d) 24 m.u. apart?

In a haploid organism, the C and D loci are 8 m.u. apart. From a cross C d × c D, give the proportion of each of the following progeny classes: (a) C D; (b) c d; (c) C d; (d) all recombinants combined.

A fruit fly of genotype B R/b r is testcrossed with b r/b r. In 84 percent of the meioses, there are no chiasmata between the linked genes; in 16 percent of the meioses, there is one chiasma between the genes. What proportion of the progeny will be B r/b r?

A three-point testcross was made in corn. The results and a recombination analysis are shown in the display below, which is typical of three-point testcrosses (p = purple leaves, + = green; v = virus-resistant seedlings, + = sensitive; b = brown midriff to seed, + = plain). Study the display, and answer parts a through c.

You have a Drosophila line that is homozygous for autosomal recessive alleles a, b, and c, linked in that order. You cross females of this line with males homozygous for the corresponding wild-type alleles. You then cross the F₁ heterozygous males with their heterozygous sisters. You obtain the following F₂ phenotypes (where letters denote recessive phenotypes and pluses denote wild-type phenotypes): 1364 + + +, 365 a b c, 87 a b +, 84 + + c, 47 a + +, 44 + b c, 5 a + c, and 4 + b +.

R. A. Emerson crossed two different pure-breeding lines of corn and obtained a phenotypically wild-type F₁ that was heterozygous for three alleles that determine recessive phenotypes: an determines anther; br, brachytic; and f, fine. He testcrossed the F₁ with a tester that was homozygous recessive for the three genes and obtained these progeny phenotypes: 355 anther; 339 brachytic, fine; 88 completely wild type; 55 anther, brachytic, fine; 21 fine; 17 anther, brachytic; 2 brachytic; 2 anther, fine.

Chromosome 3 of corn carries three loci (b for plant-color booster, v for virescent, and lg for liguleless). A testcross of triple recessives with F₁ plants heterozygous for the three genes yields progeny having the following genotypes: 305 + v lg, 275 b + +, 128 b + lg, 112 + v +, 74 + + lg, 66 b v +, 22 + + +, and 18 b v lg. Give the gene sequence on the chromosome, the map distances between genes, and the coefficient of coincidence.

Groodies are useful (but fictional) haploid organisms that are pure genetic tools. A wild-type groody has a fat body, a long tail, and flagella. Mutant lines are known that have thin bodies, are tailless, or do not have flagella. Groodies can mate with one another (although they are so shy that we do not know how) and produce recombinants. A wild-type groody mates with a thin-bodied groody lacking both tail and flagella. The 1000 baby groodies produced are classified as shown in the illustration here. Assign genotypes, and map the three genes. (Problem 25 is from Burton S. Guttman.)

164

In Drosophila, the allele dp+ determines long wings and dp determines short (“dumpy”) wings. At a separate locus, e⁺ determines gray body and e determines ebony body. Both loci are autosomal. The following crosses were made, starting with pure-breeding parents:

Use the χ² test to determine if these loci are linked. In doing so, indicate (a) the hypothesis, (b) calculation of χ², (c) p value, (d) what the p value means, (e) your conclusion, (f) the inferred chromosomal constitutions of parents, F_1, tester, and progeny.

The mother of a family with 10 children has blood type Rh⁺. She also has a very rare condition (elliptocytosis, phenotype E) that causes red blood cells to be oval rather than round in shape but that produces no adverse clinical effects. The father is Rh⁻ (lacks the Rh⁺ antigen) and has normal red blood cells (phenotype e). The children are 1 Rh⁺ e, 4 Rh⁺ E, and 5 Rh⁻ e. Information is available on the mother’s parents, who are Rh⁺ E and Rh⁻ e. One of the 10 children (who is Rh⁺ E) marries someone who is Rh⁺ e, and they have an Rh⁺ E child.

From several crosses of the general type A/A · B/B × a/a · b/b, the F₁ individuals of type A/a · B/b were testcrossed with a/a · b/b. The results are as follows:

For each set of progeny, use the χ² test to decide if there is evidence of linkage.

In the two pedigrees diagrammed here, a vertical bar in a symbol stands for steroid sulfatase deficiency, and a horizontal bar stands for ornithine transcarbamylase deficiency.

In the accompanying pedigree, the vertical lines stand for protan color blindness, and the horizontal lines stand for deutan color blindness. These are separate conditions causing different misperceptions of colors; each is determined by a separate gene.

165

In corn, a triple heterozygote was obtained carrying the mutant alleles s (shrunken), w (white aleurone), and y (waxy endosperm), all paired with their normal wild-type alleles. This triple heterozygote was testcrossed, and the progeny contained 116 shrunken, white; 4 fully wild type; 2538 shrunken; 601 shrunken, waxy; 626 white; 2708 white, waxy; 2 shrunken, white, waxy; and 113 waxy.

In the tiny model plant Arabidopsis, the recessive allele hyg confers seed resistance to the drug hygromycin, and her, a recessive allele of a different gene, confers seed resistance to herbicide. A plant that was homozygous hyg/hyg · her/her was crossed with wild type, and the F₁ was selfed. Seeds resulting from the F₁ self were placed on petri dishes containing hygromycin and herbicide.

In a diploid organism of genotype A/a; B/b; D/d, the allele pairs are all on different chromosome pairs. The two diagrams below purport to show anaphases (“pulling apart” stages) in individual cells. State whether each drawing represents mitosis, meiosis I, or meiosis II or is impossible for this particular genotype.

The Neurospora cross al-2⁺ × al-2 is made. A linear tetrad analysis reveals that the second-division segregation frequency is 8 percent.

From the fungal cross arg-6 · al-2 × arg-6⁺ · al-2⁺, what will the spore genotypes be in unordered tetrads that are (a) parental ditypes? (b) tetratypes? (c) nonparental ditypes?

For a certain chromosomal region, the mean number of crossovers at meiosis is calculated to be two per meiosis. In that region, what proportion of meioses are predicted to have (a) no crossovers? (b) one crossover? (c) two crossovers?

A Neurospora cross was made between a strain that carried the mating-type allele A and the mutant allele arg-1 and another strain that carried the mating-type allele a and the wild-type allele for arg-1 (+). Four hundred linear octads were isolated, and they fell into the seven classes given in the table below. (For simplicity, they are shown as tetrads.)

A geneticist studies 11 different pairs of Neurospora loci by making crosses of the type a · b × a⁺ · b⁺ and then analyzing 100 linear asci from each cross. For the convenience of making a table, the geneticist organizes the data as if all 11 pairs of genes had the same designation—a and b—as shown below. For each cross, map the loci in relation to each other and to centromeres.

167

Three different crosses in Neurospora are analyzed on the basis of unordered tetrads. Each cross combines a different pair of linked genes. The results are shown in the following table:

For each cross, calculate

On Neurospora chromosome 4, the leu3 gene is just to the left of the centromere and always segregates at the first division, whereas the cys2 gene is to the right of the centromere and shows a second-division segregation frequency of 16 percent. In a cross between a leu3 strain and a cys2 strain, calculate the predicted frequencies of the following seven classes of linear tetrads where l = leu3 and c = cys2. (Ignore double and other multiple crossovers.)

A rice breeder obtained a triple heterozygote carrying the three recessive alleles for albino flowers (al), brown awns (b), and fuzzy leaves (fu), all paired with their normal wild-type alleles. This triple heterozygote was testcrossed. The progeny phenotypes were

In a fungus, a proline mutant (pro) was crossed with a histidine mutant (his). A nonlinear tetrad analysis gave the following results:

In the fungus Neurospora, a strain that is auxotrophic for thiamine (mutant allele t) was crossed with a strain that is auxotrophic for methionine (mutant allele m). Linear asci were isolated and classified into the following groups:

168

A corn geneticist wants to obtain a corn plant that has the three dominant phenotypes: anthocyanin (A), long tassels (L), and dwarf plant (D). In her collection of pure lines, the only lines that bear these alleles are AA LL dd and aa ll DD. She also has the fully recessive line aa ll dd. She decides to intercross the first two and testcross the resulting hybrid to obtain in the progeny a plant of the desired phenotype (which would have to be Aa Ll Dd in this case). She knows that the three genes are linked in the order written, that the distance between the A/a and the L/l loci is 16 m.u., and that the distance between the L/l and the D/d loci is 24 m.u.

In the model plant Arabidopsis thaliana, the following alleles were used in a cross:

The T/t and D/d loci are linked 26 m.u. apart on chromosome 1, whereas the W/w and A/a loci are linked 8 m.u. apart on chromosome 2.

A pure-breeding double-homozygous recessive trichomeless nonwaxy plant is crossed with another pure-breeding double-homozygous recessive dwarf white plant.

In corn, the cross WW ee FF × ww EE ff is made. The three loci are linked as follows:

Assume no interference.

The fungal cross + · + × c · m was made, and nonlinear (unordered) tetrads were collected. The results were

In mice, the following alleles were used in a cross:

A waltzing gray bent-tailed mouse is crossed with a nonwaltzing albino straight-tailed mouse and, over several years, the following progeny totals are obtained:

Consider the Neurospora cross +; + × f; p

It is known that the +/f locus is very close to the centromere on chromosome 7–in fact, so close that there are never any second-division segregations. It is also known that the +/p locus is on chromosome 5, at such a distance that there is usually an average of 12 percent second-division segregations. With this information, what will be the proportion of octads that are

In a haploid fungus, the genes al-2 and arg-6 are 30 m.u. apart on chromosome 1, and the genes lys-5 and met-1 are 20 m.u. apart on chromosome 6. In a cross

al-2 +; + met-1 × + arg-6; lys-5 +

what proportion of progeny would be prototrophic ++; ++?

The recessive alleles k (kidney-shaped eyes instead of wild-type round), c (cardinal-colored eyes instead of wild-type red), and e (ebony body instead of wild-type gray) identify three genes on chromosome 3 of Drosophila. Females with kidney-shaped, cardinal-colored eyes were mated with ebony males. The F₁ was wild type. When F₁ females were testcrossed with kk cc ee males, the following progeny phenotypes were obtained:

From parents of genotypes A/A · B/B and a/a · b/b, a dihybrid was produced. In a testcross of the dihybrid, the following seven progeny were obtained:

A/a · B/b, a/a · b/b, A/a · B/b, A/a · b/b, a/a · b/b, A/a · B/b, and a/a · B/b

Do these results provide convincing evidence of linkage?

Use the Haldane map function to calculate the corrected map distance in cases where the measured RF = 5%, 10%, 20%, 30%, and 40%. Sketch a graph of RF against corrected map distance, and use it to answer the question, When should one use a map function?

An individual heterozygous for four genes, A/a · B/b · C/c · D/d, is testcrossed with a/a · b/b · c/c · d/d, and 1000 progeny are classified by the gametic contribution of the heterozygous parent as follows:

An autosomal allele N in humans causes abnormalities in nails and patellae (kneecaps) called the nail–patella syndrome. Consider marriages in which one partner has the nail–patella syndrome and blood type A and the other partner has normal nails and patellae and blood type O. These marriages produce some children who have both the nail–patella syndrome and blood type A. Assume that unrelated children from this phenotypic group mature, intermarry, and have children. Four phenotypes are observed in the following percentages in this second generation:

Fully analyze these data, explaining the relative frequencies of the four phenotypes. (See pages 219-220 for the genetic basis of these blood types.)

Assume that three pairs of alleles are found in Drosophila: x⁺ and x, y⁺ and y and z⁺ and z. As shown by the symbols, each non-wild-type allele is recessive to its wild-type allele. A cross between females heterozygous at these three loci and wild-type males yields progeny having the following genotypes: 1010 x⁺ · y ⁺ · z⁺ females, 430 x · y⁺ · z males, 441 x⁺ · y · z⁺ males, 39 x · y · z males, 32 x⁺ · y⁺ · z males, 30 x ⁺ · y + · z ⁺ males, 27 x· y · z ⁺ males, 1 x ⁺ · y · z male, and 0 x · y ⁺ · z ⁺ males.

170

The five sets of data given in the following table represent the results of testcrosses using parents with the same alleles but in different combinations. Determine the order of genes by inspection—that is, without calculating recombination values. Recessive phenotypes are symbolized by lowercase letters and dominant phenotypes by pluses.

From the phenotype data given in the following table for two 3-point testcrosses for (1) a, b, and c and (2) b, c, and d, determine the sequence of the four genes a, b, c, and d and the three map distances between them. Recessive phenotypes are symbolized by lowercase letters and dominant phenotypes by pluses.

Vulcans have pointed ears (determined by allele P), absent adrenals (determined by A), and a right-sided heart (determined by R). All these alleles are dominant to normal Earth alleles: rounded ears (p), present adrenals (a), and a left-sided heart (r). The three loci are autosomal and linked as shown in this linkage map:

Mr. Spock, first officer of the starship Enterprise, has a Vulcan father and an Earthling mother. If Mr. Spock marries an Earth woman and there is no (genetic) interference, what proportion of their children will have

In a certain diploid plant, the three loci A, B, and C are linked as follows:

One plant is available to you (call it the parental plant). It has the constitution A b c/a B C.

The following pedigree shows a family with two rare abnormal phenotypes: blue sclerotic (a brittle-bone defect), represented by a black-bordered symbol, and hemophilia, represented by a black center in a symbol. Members represented by completely black symbols have both disorders. The numbers in some symbols are the numbers of individuals with those types.

171

The human genes for color blindness and for hemophilia are both on the X chromosome, and they show a recombinant frequency of about 10 percent. The linkage of a pathological gene to a relatively harmless one can be used for genetic prognosis. Shown here is part of a bigger pedigree. Blackened symbols indicate that the subjects had hemophilia, and crosses indicate color blindness. What information could be given to women III-4 and III-5 about the likelihood of their having sons with hemophilia?

(Problem 63 is adapted from J. F. Crow, Genetics Notes: An Introduction to Genetics. Burgess, 1983.)

A geneticist mapping the genes A, B, C, D, and E makes two 3-point testcrosses. The first cross of pure lines is

A/A · B/B · C/C · D/D · E/E × a/a · b/b · C/C · d/d · E/E

The geneticist crosses the F₁ with a recessive tester and classifies the progeny by the gametic contribution of the F₁:

The second cross of pure lines is A/A · B/B · C/C · D/D · E/E × a/a · B/B · c/c · D/D · e/e.

The geneticist crosses the F₁ from this cross with a recessive tester and obtains

The geneticist also knows that genes D and E assort independently.

In the plant Arabidopsis, the loci for pod length (L, long; l, short) and fruit hairs (H, hairy; h, smooth) are linked 16 m.u. apart on the same chromosome. The following crosses were made:

If the F₁’s from cross i and cross ii are crossed,

In corn (Zea mays), the genetic map of part of chromosome 4 is as follows, where w, s, and e represent recessive mutant alleles affecting the color and shape of the pollen:

If the following cross is made

+ + +/+ + + × w s e/w s e

and the F₁ is testcrossed with w s e/w s e, and if it is assumed that there is no interference on this region of the chromosome, what proportion of progeny will be of the following genotypes?

Every Friday night, genetics student Jean Allele, exhausted by her studies, goes to the student union’s bowling lane to relax. But, even there, she is haunted by her genetic studies. The rather modest bowling lane has only four bowling balls: two red and two blue. They are bowled at the pins and are then collected and returned down the chute in random order, coming to rest at the end stop. As the evening passes, Jean notices familiar patterns of the four balls as they come to rest at the stop. Compulsively, she counts the different patterns. What patterns did she see, what were their frequencies, and what is the relevance of this matter to genetics?

In a tetrad analysis, the linkage arrangement of the p and q loci is as follows:

Assume that

What proportions of tetrads will be of the following types? (a) M_IM_I, PD; (b) M_IM_I, NPD; (c) M_IM_II; (d)M_IIM_I T; (e) M_IIM_II, PD; (f) M_IIM_II, NPD; (g) M_IIM_II,T. (Note: Here the M pattern written first is the one that pertains to the p locus.) Hint: The easiest way to do this problem is to start by calculating the frequencies of asci with crossovers in both regions, region i, region ii, and neither region. Then determine what M_I and M_II patterns result.

For an experiment with haploid yeast, you have two different cultures. Each will grow on minimal medium to which arginine has been added, but neither will grow on minimal medium alone. (Minimal medium is inorganic salts plus sugar.) Using appropriate methods, you induce the two cultures to mate. The diploid cells then divide meiotically and form unordered tetrads. Some of the ascospores will grow on minimal medium. You classify a large number of these tetrads for the phenotypes ARG⁻ (arginine requiring) and ARG⁺ (arginine independent) and record the following data:

An RFLP analysis of two pure lines A/A · B/B and a/a · b/b showed that the former was homozygous for a long RFLP allele (l) and the latter for a short allele (s). The two were crossed to form an F₁, which was then backcrossed to the second pure line. A thousand progeny were scored as follows: