Campbell Biology
12th Edition
ISBN: 9780135188743
Author: Urry
Publisher: PEARSON
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Chapter 14, Problem 5TYU
Flower position, stem length, and seed shape are three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as indicated in Table 14.1. If a plant that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring would you expect to be each of the following? (Note: Use the rulesof probability instead of a huge Punnett Square.)
(a) homozygous for the three dominant traits
(b) homozygous for the three recessive traits
(c) heterozygous for all three characters
(d) homozygous for axial and tall, heterozygous for seed shape
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Mendel crossed peas having round seeds and yellow cotyledons (seed leaves) with peas having wrinkled seeds and green cotyledons. All the F1 plants had round seeds with yellow cotyledons. Diagram this cross through the F2 generation, using both the Punnett square and forked-line, or branch diagram, methods.
Gray seed color in peas is dominant to white. Assume that Mendel conducted a series of experiments where plants with gray seeds were crossed with each other and the following progeny were produced: 320 gray and 80 white. (a) What is the most probable genotype of each parent? (b) Based on your answer in (a) above, what genotypic and phenotypic ratios are expected in the progeny?
Mendel testcrossed pea plants grown from yellow, round F1 seeds to plants
grown from green, wrinkled seeds and obtained the following results: 31
yellow, round; 26 green, round; 27 yellow, wrinkled; and 26 green, wrinkled.
Are these results consistent with the hypothesis that seed color and seed
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Chapter 14 Solutions
Campbell Biology
Ch. 14.1 - DRAW IT Pea plants heterozygous for flower...Ch. 14.1 - WHAT IF? List all gametes that could be made by a...Ch. 14.1 - MAKE CONNECTIONS In some pea plant crosses, the...Ch. 14.2 - Prob. 1CCCh. 14.2 - Two organisms, with genotypcs BbDD and BBDd, are...Ch. 14.2 - WHAT IF? Three characters (flower color, seed...Ch. 14.3 - What two properties, one structural and one...Ch. 14.3 - If a man with type AB blood marries a woman with...Ch. 14.3 - WHAT IF? A rooster with gray feathers and a hen...Ch. 14.4 - Prob. 1CC
Ch. 14.4 - Prob. 2CCCh. 14.4 - Prob. 3CCCh. 14.4 - MAKE CONNECTIONS In Table 14.1, note the...Ch. 14 - When Mendel did crosses of true-breeding purple-...Ch. 14 - DRAW IT Redraw the Punnett Square on The right...Ch. 14 - Inheritance patterns are often more complex than...Ch. 14 - Both members of a couple know that they are...Ch. 14 - Prob. 1TYUCh. 14 - A man with type A blood marries a woman with type...Ch. 14 - A man has six fingers on each hand and six toes on...Ch. 14 - Prob. 4TYUCh. 14 - Flower position, stem length, and seed shape are...Ch. 14 - Hemochromatosis is an inherited disease caused by...Ch. 14 - The genotype of F1, individuals in a tetrahybrid...Ch. 14 - What is the probability that each of thc following...Ch. 14 - Prob. 9TYUCh. 14 - Prob. 10TYUCh. 14 - In tigers, a recessive allele of a particular gene...Ch. 14 - In maize (com) plants,a dominant allele I inhibits...Ch. 14 - The pedigree belowtraces the inheritance of...Ch. 14 - Imagine that you are a genetic counselor, and a...Ch. 14 - EVOLUTION CONNECTION Over the past half century,...Ch. 14 - SCIENTIFIC INQUIRY You are handed a mystery pea...Ch. 14 - Prob. 17TYUCh. 14 - SYNTHESIZE YOUR KNOWLEDGE Just for fun, imagine...
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- Assume that plant height is determined by a pair of alleles at each of two loci (A and a, B and b) that are additive in their effects. Further assume that each allele represented by an uppercase letter contributes 6 feet to height and that each allele represented by a lowercase letter contributes 2 foot to height. If a plant with genotype Aabb is crossed to a plant with genotype aaBb, what heights are expected in the progeny? 1/216ft : 1/2 12ft 1/4 16ft : 1/2 12ft : 1/4 8ft 1/4 20ft : 1/2 16ft : 1/4 12ft 1/4 24ft : 1/2 20ft : 1/4 16ft 1/2 20ft : 1/2 16ftarrow_forwardMendel crossed two Pea plants for plant height and flower color Tall plant (T) is dominant to Short Plant (t). Purple Flower (P) is dominant to white flower (p). Using the following information perform the dihybrid cross using punnett squares that will predict all possible genotypes of the offspring and list the number and description of the phenotypes of the offspring. A. One plant homozygous dominant for plant height and flower color crossed with another plant homozygous recessive for plant height and heterozygous for flower color.arrow_forwardGregor Mendel examined the inheritance of two traits in pea plants: seed coat texture and colour. Seed coat texture can be represented as S-smooth and s-wrinkled, and seed coat colour can be represented as Y-yellow and y-green. SSYY plants were crossed with ssyy plants to yield F1 pea seeds that were all smooth and all yellow. By crossing plants grown from these F1 seeds, Mendel obtained four different phenotypes of F2 seeds: • smooth and green seeds wrinkled and green seeds smooth and yellow seeds wrinkled and yellow seeds ● Use the following information to answer the next question. ● The F2 phenotypic ratio that Mendel obtained upon crossing two heterozygous smooth and yellow F1 individuals would have been: smooth and green wrinkled and green : smooth and yellow: wrinkled and yellow Record only the numeric values associated with the phenotypes. (Do not include the colons, spaces, commas, etc.)arrow_forward
- In general terms, genes found on the same chromosome are linked, and will appear to defy Mendel’s Law of Independent Assortment. This law states that alleles (Links to an external site.) for different traits (Links to an external site.) are transmitted (Links to an external site.) to offspring (Links to an external site.) independently of one another. Functionally, this means that in a dihybrid testcross, in which a heterozygote is crossed to a double homozygous recessive individual, the expected 1:1:1:1 ratio will not be obtained. Instead, lower than expected numbers of non-parentals will result, because these non-parental flies are the result of recombination during synapsis. Interestingly, and functionally important in this exercise, synapsis only occurs in female fruit flies, requiring that the heterozygote in any study of linkage must be the female. Determining the relative positions of linked genes on a chromosome can be accomplished by calculating the frequency of…arrow_forwardIn Mendel's genetic experiments many characteristics of the plants were quantified, such as their height (tall or short), flower position (axial or terminal), and seed color (colored or white). Suppose 100 plants are measured, and 37 are tall with axial flowers and colored seeds. Further, suppose a total of 26 plants have white seeds, 4 of them being short and 7 having terminal flowers. There are no short plants with terminal flowers, and the total number of short plants is 12. What is the probability a randomly selected plant will have flowers in the terminal position? Round your answer to THREE DECIMAL places. Also, double check your answer before posting. thank youarrow_forwardE. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings . Q. Is there epistasis among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic?arrow_forward
- Which of the following statements is true about the molecular basis of Mendel's second law? OThe random attachment of the sister chromatids and the split of them into different daughter cells during the second round of cell division in Meiosis ensure Mendel's second law of independent assortment of alleles. OThe random attachment of the sister chromatids and the split of them into different daughter cells during the first round of cell division in Meiosis ensure Mendel's 2nd law. O The molecular basis of independent assortment of alleles is the segregation of sister chromatids during the second round of cell division in Meiosis. O The molecular basis of independent assortment of alleles is the segregation of homologous chromosomes during the second round of cell division in Meiosis. O The molecular basis of independent assortment of alleles is the segregation of homologous chromosomes during the first round of cell division in Meiosis.arrow_forwardOn the basis of Mendel’s observations (i.e. Mendelian patterns), predict the results from the following crosses with peas. Show the results in terms of genotypes, phenotypes, and their proportions in the offspring. The progeny of a tall (dominant and homozygous) variety crossed with a dwarf variety. The progeny of (a) self-fertilized. The progeny from (a) crossed with the original tall parent. The progeny of (a) crossed with the original dwarf parent.arrow_forwardMendel crossed a wrinkle-seeded plant with pure round-seeded plant. Round is the dominant trait. Create a Punnett Square to represent this relationship. Using this information outlined in this Punnett Square, what percentage of the offspring do we anticipate will be wrinkle-seeded? (Chp 2) O 100% O 75% O 50% O 25% 0 0%arrow_forward
- Mendel's concept of dominance states that in a genotype where two different alleles of a locus are present, only the trait encoded by the dominant allele is observed. Give a molecular explanation for dominance, i.e. explain intracellular molecular events that can result in what we observe as dominance on a phenotypic level. Use the gene that encodes seed shape in peas as an example, where roun(R) is dominant over wrinkled(r), to explain how RR and Rr plants can have the same phenotype.arrow_forwardIn a cross involving polygenic inheritance, three (3) gene pairs control plant height. The shortest andtallest plants are 12 cm and 24 cm, respectively. What height should all F1s display if the shortest andtallest plants were crossed, assuming environmental factors are the same?arrow_forwardA cross is performed between a pure-breeding pea plant with round and yellow seeds and a pure-breeding pea plant with wrinkled and green seeds. All F1 progeny have round and yellow seeds. The F1 plants are self-crossed. The following results are observed in the resultant F2 generation: 65 round and yellow, 42 round and green, 50 wrinkled and yellow, and 45 wrinkled and green. Use a Chi Square analysis to determine whether these traits follow Mendel’s law of independent assortment. Show all calculationsarrow_forward
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