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 4, Problem 2NST
In some plants a red flower pigment, cyanidin, is synthesized from a colorless precursor. The addition of a hydroxyl group (OH−) to the cyanidin molecule causes it to become purple. In a cross between two randomly selected purple varieties, the following results were obtained:
94 purple
31 red
43 white
How many genes are involved in the determination of these flower colors? Which genotypic combinations produce which
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In some plants a red pigment is synthesized from a colorless precursor. The addition of a hydroxyl group to the precursor molecule causes it to become purple. In a cross between two randomly selected purple plants, the following results were obtained. 94 purple 31 red 43 white
Q:What are the genotypes of the two random purple plants?
R/r; P/p X R/r; P/p
R/r; p/p X R/r; p/p
r/r; P/p X r/r; P/p
r/r; p/p X r/r; p/p
R/R; P/P X r/r; p/p
In sweet peas, the given enzymatic pathway result to pigment formation in the flowers:
A dihybrid plant is crossed to a white one which is heterozygous at the “C” locus (provided in the image).
What is the genotype of the dihybrid plant?
What is the genotype of the white plant?
What kinds of flowers, colored or white, are to be expected from the cross above? Include the ratio.
The allele G for yellow stigma is completely dominant to green (g). Supposing
two strains of autotetraploid plants are available and their genotypes are as follows:
GGgg – in this plant the gene is close to the centromere
Gggg – in this plant the gene is far from the centromere
If these two plants are crossed:
a) provide the gametes that can be obtained from the two plants;
b) provide the genotypic and phenotypic ratios of the offspring.
Chapter 4 Solutions
Concepts of Genetics (12th Edition)
Ch. 4 - In the guinea pig, one locus involved in the...Ch. 4 - In some plants a red flower pigment, cyanidin, is...Ch. 4 - Below are three pedigrees. For each trait,...Ch. 4 - Researching their family histories, a deaf couple...Ch. 4 - Researching their family histories, a deaf couple...Ch. 4 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 4 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 4 - In shorthorn cattle, coat color may be red, white,...Ch. 4 - In foxes, two alleles of a single gene, P and p,...Ch. 4 - In mice, a short-tailed mutant was discovered....
Ch. 4 - List all possible genotypes for the A, B, AB, and...Ch. 4 - With regard to the ABO blood types in humans,...Ch. 4 - In a disputed parentage case, the child is blood...Ch. 4 - The A and B antigens in humans may be found in...Ch. 4 - In chickens, a condition referred to as creeper...Ch. 4 - In rabbits, a series of multiple alleles controls...Ch. 4 - Three gene pairs located on separate autosomes...Ch. 4 - As in Problem 12, flower color may be red, white,...Ch. 4 - Horses can be cremello (a light cream color),...Ch. 4 - With reference to the eye color phenotypes...Ch. 4 - Pigment in mouse fur is only produced when the C...Ch. 4 - In rats, the following genotypes of two...Ch. 4 - Given the inheritance pattern of coat color in...Ch. 4 - In a species of the cat family, eye color can be...Ch. 4 - In a plant, a tall variety was crossed with a...Ch. 4 - In a unique species of plants, flowers may be...Ch. 4 - Five human matings (15), identified by both...Ch. 4 - A husband and wife have normal vision, although...Ch. 4 - In humans, the ABO blood type is under the control...Ch. 4 - In Drosophila, an X-linked recessive mutation,...Ch. 4 - Another recessive mutation in Drosophila, ebony...Ch. 4 - In Drosophila, the X-linked recessive mutation...Ch. 4 - While vermilion is X-linked in Drosophila and...Ch. 4 - In a cross in Drosophila involving the X-linked...Ch. 4 - Consider the three pedigrees below, all involving...Ch. 4 - In goats, the development of the beard is due to a...Ch. 4 - Predict the F1 and F2 results of crossing a male...Ch. 4 - Two mothers give birth to sons at the same time at...Ch. 4 - Discuss the topic of phenotypic expression and the...Ch. 4 - Prob. 35PDQCh. 4 - Labrador retrievers may be black, brown...Ch. 4 - A true-breeding purple-leafed plant isolated from...Ch. 4 - In Dexter and Kerry cattle, animals may be polled...Ch. 4 - A geneticist from an alien planet that prohibits...Ch. 4 - The following pedigree is characteristic of an...Ch. 4 - Students taking a genetics exam were expected to...Ch. 4 - In four oclock plants, many flower colors are...Ch. 4 - Below is a partial pedigree of hemophilia in the...
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- There are some members of the genus Datura that are trisomic for the chromosome set, which carries the genes for purple (p+) and white (p) flower color. In a cross involving a trisomic p+ p+ p female and pollen from a normal p+ p plant, what is the expected proportion of purple and white flowered individuals?arrow_forwardIn some plants, a true-breeding, red-flowered variety gives all pink flowers when crossed with a white-flowered strain: RR (red) X rr (white) = Rr (pink). Flower position in this plant may either be axial (dominant) or terminal (recessive). What will be the phenotypic ratios of the F1 generation resulting from the following cross: axial-red (true-breeding) X terminal-white? What will the phenotypic ratios of the F2 be? SHOW ALL WORKarrow_forwardIn corn, the allele f ′ causes floury endosperm and the allele f ″ causes flinty endosperm. In the cross f ′/f ′ ×f ″/f ″, all the progeny endosperms are floury, but, in the reciprocal cross, all the progeny endosperms are flinty. What is a possible explanation? (Check the legend for Figure 2-7.)arrow_forward
- 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.arrow_forwardA purple flowering, smooth seed dihybrid plant (genotype PpSs) is test crossed with a white flowering, wrinkled seed plant (genotype ppss). These produce progeny in the following numbers of four phenotypes: 24:76:74:26 (purple flower + smooth seed coat: purple flower + wrinkled seed coat: white flower + smooth seed coat: white flower + wrinkled seed coat). a) What is the genotype of the original dihybrid plant? Specify which alleles are on each chromosome of the purple flowering, smooth seed plant (i.e. AB/ab or Ab/aB). b) How many map units separate the colour and seed coat genes? Show your calculations.arrow_forwardThree autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progenyarrow_forward
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