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
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 25, Problem 7PDQ
Erma and Harvey were a compatible barnyard pair, but a curious sight. Harvey’s tail was only 6 cm long, while Erma’s was 30 cm. Their F1 piglet offspring all grew tails that were 18 cm. When inbred, an F2 generation resulted in many piglets (Erma and Harvey’s grandpigs), whose tails ranged in 4-cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, and 30). Most had 18-cm tails, while 1/64 had 6-cm tails and 1/64 had 30-cm tails.
- (a) Explain how these tail lengths were inherited by describing the mode of inheritance, indicating how many gene pairs were at work, and designating the genotypes of Harvey, Erma, and their 18-cm-tail offspring.
- (b) If one of the 18-cm-tail F1 pigs is mated with one of the 6-cm-tail F2 pigs, what
phenotypic ratio will be predicted if many offspring resulted? Diagram the cross.
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
Erma and Harvey were a compatible barnyard pair, but a curioussight. Harvey’s tail was only 6 cm long, while Erma’s was 30 cm.Their F1 piglet offspring all grew tails that were 18 cm. Wheninbred, an F2 generation resulted in many piglets (Erma andHarvey’s grandpigs), whose tails ranged in 4-cm intervals from6 to 30 cm (6, 10, 14, 18, 22, 26, and 30). Most had 18-cm tails,while 1/64 had 6-cm tails and 1/64 had 30-cm tails.
Que: If one of the 18-cm F1 pigs is mated with one of the 6-cmF2 pigs, what phenotypic ratio would be predicted if manyoffspring resulted? Diagram the cross.
pure-breeding yellow (Y) canaries crossed with pure breeding blue (y) canaries make green cancaries. identify the genotypic and phenotypic ratios when you cross a green with a blue. what type of inheritence pattern is this?
Erma and Harvey were a compatible barnyard pair, but a curioussight. Harvey’s tail was only 6 cm long, while Erma’s was 30 cm.Their F1 piglet offspring all grew tails that were 18 cm. Wheninbred, an F2 generation resulted in many piglets (Erma andHarvey’s grandpigs), whose tails ranged in 4-cm intervals from6 to 30 cm (6, 10, 14, 18, 22, 26, and 30). Most had 18-cm tails,while 1/64 had 6-cm tails and 1/64 had 30-cm tails.
Que: Explain how these tail lengths were inherited by describingthe mode of inheritance, indicating how many gene pairswere at work, and designating the genotypes of Harvey,Erma, and their 18-cm-tail offspring.
Chapter 25 Solutions
Concepts of Genetics (12th Edition)
Ch. 25 - A homozygous plant with 20-cm-diameter flowers is...Ch. 25 - The following table shows measurements for fiber...Ch. 25 - The following cable gives the percentage of twin...Ch. 25 - At an interview with a genetic counselor, a couple...Ch. 25 - Prob. 2CSCh. 25 - At an interview with a genetic counselor, a couple...Ch. 25 - HOW DO WE KNOW? In this chapter, we focused on a...Ch. 25 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 25 - Define the following: (a) polygenic, (b) additive...Ch. 25 - A dark-red strain and a white strain of wheat are...
Ch. 25 - Height in humans depends on the additive action of...Ch. 25 - An inbred strain of plants has a mean height of 24...Ch. 25 - Erma and Harvey were a compatible barnyard pair,...Ch. 25 - In the following table, average differences of...Ch. 25 - What kind of heritability estimates (broad sense...Ch. 25 - List as many human traits as you can that are...Ch. 25 - Corn plants from a test plot are measured, and the...Ch. 25 - The following variances were calculated for two...Ch. 25 - The mean and variance of plant height of two...Ch. 25 - Prob. 14PDQCh. 25 - Prob. 15PDQCh. 25 - In an assessment of learning in Drosophila, flies...Ch. 25 - Prob. 17PDQCh. 25 - Prob. 18PDQCh. 25 - In a population of 100 inbred, genotypically...Ch. 25 - Many traits of economic or medical significance...Ch. 25 - A 3-inch plant was crossed with a 15-inch plant,...Ch. 25 - In a cross between a strain of large guinea pigs...Ch. 25 - Type A1B brachydactyly (short middle phalanges) is...Ch. 25 - In a series of crosses between two true-breeding...Ch. 25 - Students in a genetics laboratory began an...Ch. 25 - Prob. 26ESPCh. 25 - Canine hip dysplasia is a quantitative trait that...Ch. 25 - Floral traits in plants often play key roles in...Ch. 25 - In 1988, Horst Wilkens investigated blind...
Knowledge Booster
Learn more about
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
- Complete a Punnett square for a genetic cross of two truebreeding Portuguese water dogs—one with a black, wavy coat (homozygous dominant, BBWW) and one with a brown, curly coat (homozygous recessive, bbww). What is the phenotype ratio of their offspring (F1)? Now fill out another Punnett square, crossing two of the offspring. What is the phenotype ratio of the F2 generation?arrow_forwardA guinea pig that is heterozygous for fur color and texture mates with another guinea pig heterozygous for both traits. They produce 96 offspring. How do I draw a Punnett Square for this information?arrow_forwardIn a breeding experiment, two heterozygous purple heterozygous tall plants are crossed. Solve for the phenotypic and genotypic ratios of F1. White color and shortness are the recessive alleles.arrow_forward
- Two pure-breeding lines of petunia plants are crossed. Line 1 plants grow to a height of 54 cm, and Line 2 plants grow to a height of 18 cm. Petunia plant height is controlled by three genes, A, B and C. Line 1 has the genotype A₁A₁B₁B₁C₁C₁, and line 2 has the genotype A2A2B₂B₂C₂C₂. Assume that genotype alone determines plant height under ideal growth conditions and that the alleles of the three genes are additive. If the F1 plants are self crossed, what is the expected proportion of F2 plants with the genotype A₁A₁B₁B₁C₁C₁ 1/8 1/32 1/16 1/4 1/64arrow_forwardPure-breeding yellow (Y) canaries (a canary is a type of bird)crossed with pure-breeding blue (y) canaries make green canaries. Identify the genotypic and phenotypic ratios when you cross a green canary with a blue canary.arrow_forwardGuinea pigs have coats that are either rough or smooth. Rough coats are dominant, so they are represented by (R), and smooth coats are recessive, so they are represented by (r). If we crossed a homozygous rough-coated guinea pig with a smooth-coated guinea pig, what are the phenotypic and genotypic ratios?arrow_forward
- A number of true-breeding brown dogs were crossed with a true-breeding white breed. All the pups were brown. When several of the F1's were allowed to breed, the F2's were 92 brown, 58 black and 10 white. Provide a genetic explanation for these results and show the genotypes of the two original parents.arrow_forwardTwo true-breeding varieties of maize, one 11 cm high and the other 47 cm high were crossed and the resultant F1 hybrids were then crossed to generate the F2 . In the F2 there were a total of 13,923 plants with a continuous variation in heights between the two extremes and with only 3 plants as large as 47 cm high and 5 plants of 11 cm high. The F1 from (b) is then crossed to give an F2. What proportion would you expect in the F2 of each extreme phenotype?arrow_forwardCorn breeders obtained pure lines whose kernels turn sun red, pink, scarlet, or orange when exposed to sunlight (normal kernels remain yellow in sunlight). Some crosses between these lines produced the following results. The phenotypes are abbreviated O, orange; P, pink; Sc, scarlet; and SR, sun red.PhenotypesCross Parents F1 F21 SR × P all SR 66 SR: 20 P2 O × SR all SR 998 SR: 314 O3 O × P all O 1300 O: 429 P4 O × Sc all Y 182 Y:80 O:58 ScAnalyze the results of each cross, and provide a unifying hypothesis to account for all the results. (Explain all symbols that you use.)arrow_forward
- In mangoes wide leaf and narrow leaf are alternative characters and wide (W) is dominant to narrow (w). Red stem and green stem are another pair of characters with red (R) dominant to green (r) . A true breeding wide, green mango plant was crossed with a true-breeding narrow, red plant and the F1 plants were interbred. 337 F2 plants were produced with the following phenotypes; 202 wide, red; 69 wide, green; 15 narrow, green; 51 narrow, red. How many degrees of freedom (df) are there in this experiment?arrow_forwardIn pea plants yellow seed color, (GG) and round seed shape (WW) seeds are dominant traits, while green color or (gg) wrinkled shape (ww) seeds are recessive traits. You cross a pure breeding plant with yellow wrinkled seeds to a pure breeding plant with green round seeds to generate F1s. When you perform a reciprocal cross you get the same results. 9) You take an F1 male and an F1 female to generate F2 offspring. You take an F2 male with round and yellow seeds and mate them with an F2 female that has round green seeds. If these F2s produce 108 offspring, how many would be predicted to be green and wrinkled? a) 27 b) 3 c) 4 d) 7 e) 12arrow_forwardIn mangoes wide leaf and narrow leaf are alternative characters and wide (W) is dominant to narrow (w). Red stem and green stem are another pair of characters with red (R) dominant to green (r) . A true breeding wide, green mango plant was crossed with a true-breeding narrow, red plant and the F1 plants were interbred. 337 F2 plants were produced with the following phenotypes; 202 wide, red; 69 wide, green; 15 narrow, green; 51 narrow, red. What probability is associated with the Chi Sq value you obtain?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Concepts of Biology
Biology
ISBN:9781938168116
Author:Samantha Fowler, Rebecca Roush, James Wise
Publisher:OpenStax College
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY