Biology: Life on Earth
10th Edition
ISBN: 9780321729712
Author: Gerald Audesirk, Teresa Audesirk, Bruce E. Byers
Publisher: Benjamin Cummings
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Chapter 10, Problem 5GP
In the tomatoes of Problem 4, an F1 offspring (RrSs) was mated with a homozygous recessive (rrss). The following offspring were obtained:
Round, smooth: 583 Long, fuzzy: 602
Round, fuzzy: 21 Long, smooth: 16
What is the most likely explanation for this distribution of
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Students have asked these similar questions
In Drosophila, a fully heterozygous female with the X-linked recessive genes a, b, and c (not necessarily in that order on
the chromosome) was mated to a male that was genetically a, b, c (not necessarily in that order on the chromosome).
The offspring occurred in the following phenotypic ratios:
Phenotypes:
Numbers:
What is the cis/trans arrangement in the heterozygous parent?
Wild
426
а, с, b
428
Which gene is in the middle?
a
23
c, b
22
If you added 23, 22, 3, and 2, it would give you the map distance between
genes
C
49
b, a
46
What calculation would you make to determine if interference was
occurring? (you don't have to complete the calculation)
b.
C, a
Total
1000
3.
In an insect species, the genes for white eyes (w) and smooth legs (s) are linked and are located 16
map units apart.
A white-eyed, smooth-legged female was mated to a true-breeding wild-type male; the resulting F1
phenotypically wild-type females were mated to white-eyed, smooth-legged males.
In an F2 of 100 individuals, what would be the expected number of F2 individuals for each
phenotype?
wild-type:
white-eyed:
smooth-legged:
white-eyed and smooth-legged:
A wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wildtype, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?
Chapter 10 Solutions
Biology: Life on Earth
Ch. 10 - Fill-in-the-Blank The physical position of a gene...Ch. 10 - Define nondisjunction, and describe common...Ch. 10 - In certain cattle, hair color can be red...Ch. 10 - Prob. 2GPCh. 10 - In the edible pea, tall (T) is dominant to short...Ch. 10 - In tomatoes, round fruit (R) is dominant to long...Ch. 10 - In the tomatoes of Problem 4, an F1 offspring...Ch. 10 - Prob. 6GPCh. 10 - In humans, one of the genes determining color...Ch. 10 - In the couple described in Problem 7, the woman...
Ch. 10 - An organism is described as Rr, with red coloring....Ch. 10 - 2. The inheritance of multiple traits depends on...Ch. 10 - Fill-in-the-Blank Many organisms, including...Ch. 10 - 4. Genes that are present on one sex chromosome...Ch. 10 - 5. When the phenotype of heterozygotes is...Ch. 10 - 1. Define the following terms: gene, allele,...Ch. 10 - 2. Explain why genes located on the same...Ch. 10 - 3. Define polygenic inheritance. Why does...Ch. 10 - What is sex linkage? In mammals, which sex would...Ch. 10 - What is the difference between a phenotype and a...Ch. 10 - 6. In the pedigree of part (a) of Figure 11–22, do...Ch. 10 - Sometimes the term gene is used rather casually....Ch. 10 - In an alternate universe, all the genes in all...Ch. 10 - Prob. 3AC
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- Females from a pure-breeding curly-winged strain are mated with males from a pure-breeding straight-winged (wild-type) strain. The F1 mate with each other to produce an F2 generation that consists of 160 flies with curly wings and 80 with straight wings. What can you infer from this observation? Explain your answer and why the other options do not qualify, A) Curly wings is a recessive trait. B) The dominant curly wing allele is also a recessive lethal. C) Wing shape is controlled by two codominant alleles. D) Two interacting genes determine wing shape. E) All of the hybrid F 1 flies had straight wings.arrow_forwardIn a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the gene order of these linked genes?arrow_forwardIn the video game Animal Crossing: New Horizons, flowering breeding is based in genetics. Each flower's color is determined by the genotype at three or four unlinked genes: R, Y, W, and S. The genotype of the elusive blue rose is RR YY ww ss. In the game, one way to get a blue rose is to cross two roses with the Rr Yy Ww ss genotype. A) What types of gametes and in what proportions will a Rr Yy Ww ss rose produce? B) In a cross Rr Yy Ww ss x Rr Yy Ww ss what are the possible offspring genotypes and at what frequency will they each appear? Show your work. C) What proportion of the offspring of the cross will be blue roses?arrow_forward
- Phenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed to homozygous light eyed, straw bristled males: Phenotype Number light-straw 140 wild-type 160 light 360 straw 340 Total 1000 Compute the map distance between the light and straw loci. Group of answer choices 70 map units 3 map units 7 map units 0.03 map units 30 map unitsarrow_forward1) In some species of pea plant, pea colour is determined by one pair of alleles and pea shape by another - the genes are on different chromosomes. A plant grown from a round green pea was self-pollinated. The following numbers of offspring were obtained: 95 green round, 28 white round, 30 green wrinkled, 11 white wrinkled. a) What is the ratio of the phenotypes in the offspring? b) In each pair of alleles, colour and shape, which allele is dominant and which recessive? c) what was the genotype of the parent(s)? Use a diagram to show how this explains the phenotype ratio of the offspring.arrow_forwardIn pea plants, a dihybrid for the recessive a and b is testerossed. The distribution of the phenotypes is as follows: Phenotype AB Number 138 a b 132 a B 69 A b 61 (i) Are the genes assorting independently? Test the hypothesis with a chi-square test. State your hypothesis, estimate the p value using the x table below and clearly state your conclusion. Table 1: Chi-square table Chi Square Values and Probability Degrees of Freedom P = 0.99 0.95 0.80 0.50 0.20 0.05 0.01 0.000157 0.00393 0.0642 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3. 0.115 0.352 1.005 2.366 4.642 7.815 11.345 0.297 0.711 1.649 3.357 5.989 9.488 13.277 (ii) If the genes are linked, calculate the map distance between the genes in cM. '.arrow_forward
- A researcher crosses mice with apricot eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes:F1: 21 apricot, short : 30 brown, long : 19 brown, short : 110 apricot, longYou know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these results.a) What is your null hypothesis for the chi-square test? b) You obtain a value of 11.33 for the chi-square test. What conclusion can you make from the results of the chi-square test? c) Give the expected number of individuals in the “apricot, long” class. d) Making use of the appropriate genetic convention for naming alleles, give the genotype of any of the progeny in the “apricot, long” class.arrow_forwardA researcher crosses mice with brown eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 6 apricot, short : 30 brown, long : 15 brown, short : 9 apricot, long You know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these results. a) Making use of the appropriate genetic convention for naming alleles, give the genotype of the male parent in this cross. b) What is your null hypothesis for the chi-square test? c) Give the expected number of individuals in the "brown, long" class. d) You obtain a value of 3.47 for the chi-square test. What conclusion can you make from the results of the chi-square test? P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357…arrow_forwardA female from true breeding line of Drosophila with white eyes is crossed with a male from a true breeding line with brown eyes. All of the offspring have wild type brick red eyes. Which of the following explanations is most likely? A) There are many alleles for the single gene for eye color. Wild type brick red eyes result only when the fly is heterozygous. B) The alleles for brown, white, and brick red eyes are alleles for a single locus. The allele for brown eye color is dominant to the allele for brick red eye color and to the allele for white eyes. C) There is more than one gene for eye color. The brown mutation and the white mutation occur in separate genes and are both recessive to the wild type alleles. The offspring are heterozygous for both genes, so they are phenotypically wild type. D) None of the above. It is not possible for a cross between a white-eyed and a brown-eyed fly to produce wild type offspring.arrow_forward
- Still referring to Problem 1, what will be the possible genotypes of offspring from the following matings? With what frequency will each genotype show up? a. AABB aaBB b. AaBB AABb c. AaBb aabb d. AaBb AaBbarrow_forwardA pure breeding strain of squash that produced disk-shaped fruits was crossed with a pure- breeding strain having long fruits. The first filial generation had disk fruits, but the second filial generation showed a new phenotype, sphere, and was composed of the following proportions: disk 270, sphere 178, long 32. Propose an explanation for these results, and show the genotypes of P, First filial generation and second filial generation.arrow_forwardAs stated in Solved Problem 2, recessive mutation in certain mice called waltzers causes them to execute bizarre steps. W. H. Gates crossed waltzers with pure-breeding normal mice and found, among several hundred normal progeny, a single waltzing female mouse. This mouse was mated with a waltzing male, and her offspring were waltzers. When mated with a homozygous normal male, all her progeny were normal. Some of these normal males and females were intercrossed, and, unexpectedly, none of their progeny were waltzers. T. S. Painter examined the chromosomes of some of Gates’s waltzing mice that showed a breeding behavior similar to that of the original, unusual waltzing female. He found that these mice had the normal number of 40 chromosomes. In the unusual waltzers, however, one member of a chromosome pair was abnormally short. Interpret these observations as completely as possible, both genetically and cytologically.arrow_forward
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