EBK CAMPBELL BIOLOGY IN FOCUS
2nd Edition
ISBN: 8220101459299
Author: Reece
Publisher: PEARSON
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Chapter 11, 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 11.1. 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 as follows? (Note: Use the rules of 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.
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
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1/4 24ft : 1/2 20ft : 1/4 16ft
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Mendel 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.
Chapter 11 Solutions
EBK CAMPBELL BIOLOGY IN FOCUS
Ch. 11.1 - DRAW IT Pea plants heterozygous for flower...Ch. 11.1 - List all gametes that could be made by a pea plant...Ch. 11.1 - Prob. 3CCCh. 11.2 - For any gene with a dominant allele A and...Ch. 11.2 - Two organisms, with genotypes BbDD and BBDd, are...Ch. 11.2 - Prob. 3CCCh. 11.3 - Incomplete dominance and epistasis are both terms...Ch. 11.3 - Prob. 2CCCh. 11.3 - WHAT IF? A rooster with gray feathers and a hen of...Ch. 11.4 - Beth and Tom each have a sibling with cystic...
Ch. 11.4 - MAKE CONNECTIONS In Table 11.1, note the...Ch. 11 - DRAW IT Two pea plants heterozygous for the...Ch. 11 - A man with type A blood marries a woman with type...Ch. 11 - A man has six fingers on each hand and six toes on...Ch. 11 - Prob. 4TYUCh. 11 - Flower position, stem length, and seed shape are...Ch. 11 - Prob. 6TYUCh. 11 - Prob. 7TYUCh. 11 - Prob. 8TYUCh. 11 - Prob. 9TYUCh. 11 - Prob. 10TYUCh. 11 - In tigers, a recessive allele that is pleiotropic...Ch. 11 - Prob. 12TYUCh. 11 - Imagine that you are a genetic counselor, and a...Ch. 11 - Prob. 15TYUCh. 11 - Prob. 16TYUCh. 11 - Prob. 17TYU
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- 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 texture are controlled by independently assorting genes, each segregating two alleles?arrow_forwardQ6. Guinea pigs, which were homozygous for long, black hair were crossed with ones which were homozygous for short, white hair. All the F1 offspring had short, black hair. (a) Using suitable symbols, draw a genetic diagram showing parents genotypes, to explain this result in F1 offspring. (b) Complete the Punnett square to show the results of interbreeding the F1 offspring. Gametesarrow_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
- Gregor 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_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_forwardMendel crossed peas having round seeds and yellow cotyledons 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 methods.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_forwardAssume that smooth seed coats are dominant over rough seed coat. If two pea plants that are both Heterozygous for seed coats are crossed, what do you expect in the offspring (the f1 generation)? Specifically, what genotypes, what phenotypes, and in what ratios? (Use S for the smooth seed coat and s for the rough seed coat.) (Again, it is helpful to write out all the possible genotypes and give the phenotype for each genotype before answering the specific question.) Be careful in writing capital S and lower case.arrow_forwardMendel'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_forward
- In 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_forwardMendel's Law of Independent Assortment refers to [ Select ] In the picture below, what assorts independently from each other? [ Select ] If the cell in the picture below undergoes S phase and then Meiosis, what are the possible gamete types? [ Select ] Parent cell A- 2n=4 a- B.arrow_forwardMendel found that round (R) seed is dominant over wrinkled (r) seed, and yellow (G) seed is dominant over green (g) seed. The following crosses were made between plants of known phenotypes but unknown genotypes: 1) round, yellow X round, yellow 2) round, yellow X round, yellow 3) round, yellow X wrinkled, green b) Cross 1 2 Using the same symbols as defined above, list the genotypes of the parent plants in each cross. 3 round, yellow X round, yellow round, yellow X round, yellow round, yellow X wrinkled, green Which of the above crosses is a testcross? Offspring and ratio 3/4 round, yellow 1/4 wrinkled, yellow 9/16 round, yellow 3/16 round green 3/16 wrinkled, yellow 1/16 wrinkled, green MacBook Pro 1/4 round, yellow 1/4 round, green 1/4 wrinkled, yellow 1/4 wrinkled, green Focuarrow_forward
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