The result of a test cross of a yeast cell that is heterozygous for three genes are summarized in the table (below), which shows the number of offspring that inherited each combination of alleles from the heterozygous parent. Use the data collected to make a genetic map of the three genes L, Q, and D. I I I L I L Alleles L Q d L q q Q q q Q D D D d d d 104 27 78 64 28 62 106 81

The result of a test cross of a yeast cell that is heterozygous for three genes are summarized in the table (below), which shows the number of offspring that inherited each combination of alleles from the heterozygous parent. Use the data collected to make a genetic map of the three genes L, Q, and D. I I I L I L Alleles L Q d L q q Q q q Q D D D d d d 104 27 78 64 28 62 106 81

BIOLOGY:CONCEPTS+APPL.(LOOSELEAF)

10th Edition

ISBN:9781305967359

Author:STARR

Publisher:STARR

Chapter13: Patterns In Inherited Traits

Section: Chapter Questions

Problem 6GP: In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele...

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In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele for red flowers (p). An allele for long pollen grains (L) is dominant when paired with a recessive allele for round pollen grain (l). Bateson and Punnett crossed a plant having purple flowers/long pollen grains with one having white/flowers/round pollen grains. All F1 offspring had purple flowers and long pollen grains. Among the F2 generation, the researchers observed the following phenotypes: 296 purple flowers/long pollen grains 19 purple flowers/round pollen grains 27 red flowers/long pollen grains 85 red flowers/round pollen grains What is the best explanation for these results?
A researcher crosses two white-flowered lines ofAntirrhinum plants as follows and obtains the followingresults:pure line 1 × pure line 2↓F1 all whiteF1 × F1↓F2 131 white29 reda. Deduce the inheritance of these phenotypes; useclearly defined gene symbols. Give the genotypes ofthe parents, F1, and F2.b. Predict the outcome of crosses of the F1 with eachparental line.
We have dealt mainly with only two genes, but the sameprinciples hold for more than two genes. Consider thefollowing cross:A/a ; B/b ; C/c ; D/d ; E/e × a/a ; B/b ; c/c ; D/d ; e/ea. What proportion of progeny will phenotypicallyresemble (1) the first parent, (2) the second parent,(3) either parent, and (4) neither parent?b. What proportion of progeny will be genotypically thesame as (1) the first parent, (2) the second parent,(3) either parent, and (4) neither parent?Assume independent assortment.
Consider this cross in pea plants: Tt Rr yy Aa × Tt rr Yy Aa, whereT = tall, t = dwarf, R = round, r = wrinkled, Y = yellow, y = green,A = axial, a = terminal. What is the expected phenotypic outcomeof this cross? Have one group of students solve this problem bymaking one big Punnett square, and have another group solve it bymaking four single-gene Punnett squares and using the multiplication method. Time each other to see who gets done first.
In a fungus with four ascospores, a mutant allele lys-5causes the ascospores bearing that allele to be white,whereas the wild-type allele lys-5+ results in black ascospores. (Ascospores are the spores that constitute thefour products of meiosis.) Draw an ascus from each ofthe following crosses:a. lys-5 × lys-5+b. lys-5 × lys-5c. lys-5+ × lys-5+
Do a Punnett square for a cross between a homozygous red-eyed female (XR XR ) and a white-eyed male (Xr Y). What would be the genotypes of the female F1 flies? ____ phenotype? ______ What would be the genotypes of the male F1 flies? ___ phenotype? _______ Do a Punnett square for the F2 results if one of the F1 females mated with one of the F1 males. *To show results of sex-linked traits: Divide phenotypes into males and females first, keep results as fractions (or %). List the phenotypes, with the corresponding genotype (in parentheses) next to each phenotype. Males: Females: A sex-linked trait in humans: hemophilia = blood clotting disorder. Normal clotting (XN) is dominant to hemophilia (Xn). Do a Punnett square for a normal clotting (homozygous dominant) woman and a man with hemophilia.
. A geneticist mapping the genes A, B, C, D, and E makestwo 3-point testcrosses. The first cross of pure lines isA/A ⋅ B/B ⋅ C/C ⋅ D/D ⋅ E/E × a/a ⋅ b/b ⋅ C/C ⋅ d/d ⋅ E/EThe geneticist crosses the F1 with a recessive tester andclassifies the progeny by the gametic contribution ofthe F1:A ⋅ B ⋅ C ⋅ D ⋅ E 316a ⋅ b ⋅ C ⋅ d ⋅ E 314A ⋅ B ⋅ C ⋅ d ⋅ E 31a ⋅ b ⋅ C ⋅ D ⋅ E 39A ⋅ b ⋅ C ⋅ d ⋅ E 130a ⋅ B ⋅ C ⋅ D ⋅ E 140A ⋅ b ⋅ C ⋅ D ⋅ E 17a ⋅ B ⋅ C ⋅ d ⋅ E 131000The 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 arecessive tester and obtainsA ⋅ B ⋅ C ⋅ D ⋅ E 243a ⋅ B ⋅ c ⋅ D ⋅ e 237A ⋅ B ⋅ c ⋅ D ⋅ e 62a ⋅ B ⋅ C ⋅ D ⋅ E 58A ⋅ B ⋅ C ⋅ D ⋅ e 155a ⋅ B ⋅ c ⋅ D ⋅ E 165a ⋅ B ⋅ C ⋅ D ⋅ e 46A ⋅ B ⋅ c ⋅ D ⋅ E 341000The geneticist also knows that genes D and E assortindependently.a. Draw a map of these genes, showing distances inmap units wherever possible.b. Is there any evidence of interference?
When many families were tested for the ability to tastethe chemical phenylthiocarbamide, the matings weregrouped into three types and the progeny were totaled,with the results shown below:ChildrenNumber NonParents of families Tasters tastersTaster × taster 425 929 130Taster × nontaster 289 483 278Nontaster × nontaster 86 5 218With the assumption that PTC tasting is dominant (P)and nontasting is recessive (p), how can the progenyratios in each of the three types of mating be accountedfor?
Based on the ideas proposed by Morgan, which of the followingstatements concerning linkage is not true?a. Traits determined by genes located on the same chromosome arelikely to be inherited together.b. Crossing over between homologous chromosomes can create newallele combinations.c. A crossover is more likely to occur in a region between two genesthat are close together than in a region between two genes that arefarther apart.d. The probability of crossing over depends on the distance betweenthe genes.e. Genes that tend to be transmitted together are physically located onthe same chromosome
A wild-type fruit fly (heterozygous for gray body color andred eyes) is mated with a black fruit fly with purple eyes. Theoffspring are wild-type, 721; black purple, 751; gray purple, 49;black red, 45. What is the recombination frequency betweenthese genes for body color and eye color? Using informationfrom problem 3, what fruit flies (genotypes and phenotypes)would you mate to determine the order of the body color, wingsize, and eye color genes on the chromosome?
. 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.a. If the two genes are unlinked, what percentage of seeds are expected to grow?b. In fact, 13 percent of the seeds grew. Does this percentage support the hypothesis of no linkage? Explain. If not, calculate the number of map units between the loci.c. Under your hypothesis, if the F1 is testcrossed, what proportion of seeds will grow on the medium containing hygromycin and herbicide?
Fruit flies can have straight wings (S) or curly wings (s), and they can have be female XX or male XY. (A) For a standard monohybrid cross (Ss ´ Ss), what proportion of the offspring will have the genotype ss? (Express the proportion as a simple fraction) (B) For the following cross (SsXX ´ SsXY), what proportion of the offspring will have the genotype Ss? (Express the proportion as a simple fraction) (C) What proportion will have the genotype XX? (Express the proportion as a simple fraction) (D) What proportion will have the genotype SsXX? (Express the proportion as a simple fraction)SHOW YOUR WORK