Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 5, Problem 26PDQ
In a certain plant, fruit is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the following results.
Determine the location of the genes relative to one another and the genotypes of the two parental plants.
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Students have asked these similar questions
Two genes in chromosome 7 of corn are identified by the
recessive alleles gl (glossy), determining glossy leaves, and
ra (ramosa), determining branching of ears. When a plant
heterozygous for each of these alleles was crossed with a
homozygous recessive plant, the progeny consisted of the
following genotypes with the numbers of each indicated:
Calculate the frequency of recombination (as a decimal)
between these genes.
Gl ra / gl
ra
Gl Ra/gl
ra
98
7
gl Ra/gl
ra
gl ra / gl
ra
91
4
The following results were obtained from a dihybrid cross between two pea plants:
315 plants with round seeds and yellow flowers
101 plants with wrinkled seeds and yellow flowers
108 plants with round seeds and white flowers
32 plants with wrinkled seeds and white flowers
Assume that round seeds and yellow flowers are dominant and both parent plants were heterozygous for both traits.
A variety of opium poppy with lacerate leaves was crossed with a variety that has normal leaves. All the F1 plants had lacerate leaves. Two F1 plants were then interbred to produce the F2. Of the F2 plants, 249 had lacerate leaves and 16 had normal leaves. What are the genotypes of the plants in the P (parental), F1, and F2 generations? How are lacerate leaves determined in the opium poppy? Note: You can use the symbols A and B to represent the two genes involved. (How do we know there are two?)
Chapter 5 Solutions
Concepts of Genetics (12th Edition)
Ch. 5 - In a family with one autistic child the risk for...Ch. 5 - Given that the prenatal test can provide only a...Ch. 5 - Consider two hypothetical recessive autosomal...Ch. 5 - With two pairs of genes involved (P/p and Z/z), a...Ch. 5 - In Drosophila, a heterozygous female for the...Ch. 5 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 5 - Review the Chapter Concepts list on page 94. Most...Ch. 5 - Describe the cytological observation that suggests...Ch. 5 - Why does more crossing over occur between two...Ch. 5 - Explain why a 50 percent recovery of...
Ch. 5 - Why are double-crossover events expected less...Ch. 5 - What is the proposed basis for positive...Ch. 5 - What two essential criteria must be met in order...Ch. 5 - The genes dumpy (dp), clot (cl), and apterous (ap)...Ch. 5 - Colored aleurone in the kernels of com is due to...Ch. 5 - In the cross shown here, involving two linked...Ch. 5 - In a series of two-point mapping crosses involving...Ch. 5 - Two different female Drosophila were isolated,...Ch. 5 - In Drosophila, a cross was made between femalesall...Ch. 5 - Another cross in Drosophila involved the...Ch. 5 - In Drosophila, Dichaete (D) is a mutation on...Ch. 5 - Drosophila females homozygous for the third...Ch. 5 - In Drosophila, two mutations, Stubble (Sb) and...Ch. 5 - If the cross described in Problem 18 were made,...Ch. 5 - Are mitotic recombinations and sister chromatid...Ch. 5 - What possible conclusions can be drawn from the...Ch. 5 - An organism of the genotype AaBbCc was testcrossed...Ch. 5 - Based on our discussion of the potential...Ch. 5 - Traditional gene mapping has been applied...Ch. 5 - DNA markers have greatly enhanced the mapping of...Ch. 5 - In a certain plant, fruit is either red or yellow,...Ch. 5 - Two plants in a cross were each heterozygous for...Ch. 5 - A number of humanmouse somatic cell hybrid clones...Ch. 5 - A female of genotype produces 100 meiotic tetrads....Ch. 5 - In laboratory class, a genetics student was...Ch. 5 - Drosophila melanogaster has one pair of sex...Ch. 5 - In Drosophila, a female fly is heterozygous for...Ch. 5 - The gene controlling the Xg blood group alleles...Ch. 5 - Prob. 34ESP
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- A cross was performed using Drosophila melanogaster involving a female known to be heterozygous for both ebony body and sepia eyes and a male known to be homozygous wild type male. The resulting progeny were allowed to mate with one another to produce the data set. Three repetitions of the experiment were conducted. The following data were produced from the crosses. Test these data to determine if they are significantly different from the expected phenotypic ratio. Use the 5% level of significance. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject for each experiment. Wild eye Wild body – 112, Wild eye Ebony body – 40, Sepia eye Wild body – 35, Sepia eye Ebony body – 11arrow_forwardIn a certain species of plant, a dihybrid (round, pink seeds) is mated with a test-cross strain (wrinkled, white seeds). The following progeny are obtained: round, pink – 440 round, white – 76 wrinkled, pink - 84 wrinkled, white - 400 What is the distance (in map units) between the two genes in question? Your answer should be a single whole number between 0 and 50. Choose 50 if you think the genes are not linked.arrow_forwardSuppose that the wild type phenotype for tomatoes is red, round fruit with three-parted leaves and that a strain of mutant tomatoes has yellow, pear-shaped fruit with entire leaves. A and a represent the alleles for fruit color, B and b represent the alleles for fruit shape, and C and c the leaf shape. Two tomatoes with the genotypes Aa Bb cc and Aa Bb CC are crossed. The resulting Punnett square is shown. One cell has been completed with the correct tomato phenotype. Fill in the rest of the Punnett square by placing a fruit and leaf phenotype in each cell. ABC AbC aBC abC ABC Abc aBc abc Answer Bankarrow_forward
- In the Roselle plant, the dominant allele of gene 'A' codes for a pink flower and the recessive form codes for yellow flowers, while the dominant form of gene 'B' codes for serrated leaves, while the recessive form codes for smooth leaves. If the genes are trans linked, what are the expected genotypes and phenotypes and the respective ratios when plants heterozygous for both genes are test crossed?arrow_forwardList the genes found on each chromosome in the parents used in the testcross.arrow_forwardDetermine the coefficient of coincidence and the interference among these genes.arrow_forward
- Assume that three loci, each with two alleles (A and a, B and b, C and c), determine the difference in height between two homozygous strains of a plant. These genes are additive and equal in their effects on plant height. One strain (aa bb cc) is 10 cm in height. The other strain (AA BB CC) is 22 cm in height. The two strains are crossed, and the resulting F1 are interbred to produce F2 progeny. Give the phenotypes and the expected proportions of the F2 progeny.arrow_forwardA variety of opium poppy (Papaver somniferum L.) with lacerate leaves was crossed with a variety that has normal leaves. All the F1 had lacerate leaves. Two F1 plants were interbred to produce the F2. Of the F2, 249 had lacerate leaves and 16 had normal leaves. Give genotypes for all the plants in the P, F1, and F2 generations. Explain how lacerate leaves are determined in the opium poppy.arrow_forwardIn the red spotted butterfly there are two traits under investigation. The wing margin can be either notched (N) or straight (n) and the wing color can be either pale yellow (b) or bright yellow (B). A bright yellow male with straight wing margins was crossed with a bright yellow female with notched wing margins. All of the resulting progeny had notched wings and both color patterns were present. From these data determine the genotypes of the butterflies used in the original cross.arrow_forward
- Three corn seed traits are C for red, c for white; S for plump, s for shrunken; W for normal, w for waxy. Use the data from the testcross to map the distances and order of these three loci. Show calculations and illustrate the gene map. (Hint: drawing out the chromosomes and gametes will help.) Parents: CCssWW (CsW/CsW) X ccSSww (cSw/cSw) Trihybrid: CcSsWw (CsW/cSw) X ccssww (csw/csw) Test cross offspring Seed trait Gamete from trihybrid Number Red, shrunken, normal CsW 2777 White, plump, waxy cSw 2708 R ed, plump, waxy CSw 116 White, shrunken, normal csW 123 Red, shrunken, waxy Csw 643 White, plump, normal cSW 626 Red, plump, normal CSW 4 White, shrunken, waxy csw 3 Total number of progeny: 7000arrow_forwardIn the garden pea, yellow cotyledon color is dominant to green, and inflated pod shape is dominant to the constricted form. Considering both of these traits jointly in self-fertilized dihybrids, the progeny appeared in the following numbers: 193 green, inflated 184 yellow, constricted 556 yellow, inflated 61 green, constricted Do these genes assort independently? Support your answer using Chi-square analysis.arrow_forwardComplete the punnett square below for a cross between AA and aa A 21 a QUESTION 4 A Aa Complete the punnett square below for a cross between Aa and Aa. 03 Aa Aaarrow_forward
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