Campbell Biology
12th Edition
ISBN: 9780135188743
Author: Urry
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Question
Chapter 15, Problem 5TYU
Summary Introduction
To calculate: The recombination frequency of given genes.
Introduction: Test cross is defined as a method used to find out the genotype of any plant with the dominant character when it is not known whether it is homozygous or heterozygous for the trait.
Recombination frequency is used to find the linkage between the genes.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
You have determined that the gene order for three linked genes being studied is A B C. The number of recombinants resulting from crossover between genes A and BC are 44 and 48, respectively, while the double-crossover progeny total 4 and 6. What is the recombination frequency between genes A and B if the total progeny from the cross is 1000? Show all of your work.
Consider the recombination frequencies data below from a series of two-point crosses that were carried out for genes found at seven loci
(a, b, c, d, e, f, and g). Use this data to answer the following questions. The seven genes above are found on how many linkage groups?
Loci
a and b
a and c
a and d
a and e
a and f
a and g
b and c
b and d
b and e
b and f
b and g
a. 7 linkage groups
b. 5 linkage groups
c. 4 linkage groups
O d. 3 linkage groups
% Recombination
50
50
50
50
50
50
50
50
50
29
11
Loci
c and d
c and e
c and f
c and g
d and e
d and f
d and g
e and f
e and g
fand g
% Recombination
29
41
50
50
12
50
50
50
50
18
(1)
Cougar coat color and eye color are determined by two genes on the same chromosome, gene C and gene E. Gene C has two alleles: the dominant allele C produces wild type (tan) coats and the recessive allele c produces white coats. Gene E has two alleles: the dominant allele E produces wild type (brown eyes) and the recessive allele e confers blue eyes. To determine the recombination rate between these two loci you cross a wild-type cougar (CCEE) to a white-coated, blue-eyed cougar to produce a diheterozygous male cougar (CcEe). You then mate F1 male to a series of white-coated, blue-eyed female cougars to produce a panel of F2 offspring in the table below. What is the recombination rate between the two loci? Express your answer in map units, rounded to the nearest integer.
phenotype
number of F1 offspring
tan coat, brown eyes
200
tan coat, blue eyes
50
white coat, blue eyes
201
white coat, brown eyes
55
Chapter 15 Solutions
Campbell Biology
Ch. 15.1 - MAKE CONNECTIONS Review the description of...Ch. 15.1 - WHAT IF? Propose a possible reason that the first...Ch. 15.1 - Which one of Mendel's laws describes the...Ch. 15.2 - A white-eyed female Drosophila is mated with a...Ch. 15.2 - Neither Tim nor Rhoda has Duchenne muscular...Ch. 15.2 - MAKE CONNECTIONS Consider what you learned about...Ch. 15.3 - When two genes are located on the same chromosome,...Ch. 15.3 - VISUAL SKILLS For each type of offspring of the...Ch. 15.3 - Prob. 3CCCh. 15.4 - Prob. 1CC
Ch. 15.4 - Prob. 2CCCh. 15.4 - Prob. 3CCCh. 15.5 - Gene dosagethe number of copies of a gene that are...Ch. 15.5 - Reciprocal crosses between two primrose varieties,...Ch. 15.5 - WHAT IF? Mitochondrial genes are critical to the...Ch. 15 - What characteristic of the sex chromosomes allowed...Ch. 15 - Why are males affected by X-Iinked disorders much...Ch. 15 - Why are specific alleles of two distant genes more...Ch. 15 - Prob. 15.4CRCh. 15 - Explain how genomic imprinting and inheritance of...Ch. 15 - A man with hemophilia (a recessive, sex-linked...Ch. 15 - Pseudohypertrophic muscular dystrophy is an...Ch. 15 - Prob. 3TYUCh. 15 - A planet is inhabited by creatures that reproduce...Ch. 15 - Prob. 5TYUCh. 15 - A wild-type fruit fly (heterozygous for gray body...Ch. 15 - Assume that genes, A and B are on the same...Ch. 15 - Two genes of a flower, one Controlling blue (B)...Ch. 15 - Prob. 9TYUCh. 15 - Banana plants, which are triploid, are seedless...Ch. 15 - Prob. 11TYUCh. 15 - SCIENTIFIC INQUIRY DRAW IT Assume you are mapping...Ch. 15 - WRITE ABOUT A THEME: INFORMATION The continuity of...Ch. 15 - SYNTHESIZE YOUR KNOWLEDGE Butter flies have an X-Y...
Knowledge Booster
Similar questions
- A homozygous red pigmented female beetle (bb) is bred with a black pigmented male of unknown genotype. It’s unknown if the male homozygous dominant (BB) or heterozygous (Bb) for the pigment trait. The color of their offspring will help identify the genotype of the male. What do you call this method used to determine the genotype of the male beetle? Explain your answer by using Punnett square to show possible offspring genotypes.arrow_forwardIn rabbits the dominant allele B produces a black coat color and the recessive genotype bb produces a brown coat. Body fat color in rabbits is white if a dominant allele Y is present and yellow if the genotype is yy. Assume the mating between rabbits with the following genotypes : BBYY x bbyy produces BbYy Calculate the recombination frequency if the BbYy offspring are mated back to brown rabbits having yellow fat bbyy and the following ratio of phenotypes is observed: Coat color Fat color # Observed Black White 125 Black Yellow 75 Brown White 85 Brown Yellow 115arrow_forwardHemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?arrow_forward
- rdentify the statement that best interprets the results of the chi-square analysis. Refer to the chi-square distribution table to identify the statement that best interprets the chi-square results. a) It is not unusual that a heterozygous man produced 16 out of 25 offspring with heterochromia b) A total number of normal offspring is significantly lower than expected c) It is unusual that a man with heterochromia had 16 out of 25 offspring with heterochromia d) There is a significant difference between the observed number of offspring with heterochromia and the number of offspring that were expected to have heterochromia 8. Suppose two parents, a father with the genotype AaBbCcDdee and a mother with the genotype aaBbCCDdEe, want to have children. Assume each locus follows Mendelian inheritance patterns for dominance. What proportion of offspring will have each of the specified characteristics? Round your answers to two decimal places. a. Same genotype as father%3D b. Same genotype as…arrow_forwardThe flowers of 4 o'clock plants may be red, pink or white. Reds crossed to whites only produced pink offspring. When pink flowered plants were crossed with each other, they produced 113 red, 129 white, and 242 pink. It is hypothesized that these colors are controlled by a single gene pair of incompletely dominant alleles. Is this hypothesis acceptable on the basis of a Chi-square test? Show your solutions below. Chi computed value: Degree of freedom: Chi tabulated value: (at a = 0.05) Conclusion:arrow_forwardGive only typing answer with explanation and conclusion to all parts Two 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. a) How many i) genes and ii) how many alleles are involved in determining height in this plant? b) What is the contribution of each dominant allele to the phenotype in cm?arrow_forward
- In a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. Why is the recombination frequency for the outside pair of genes not equal to the sum of the recombination between the adjacent gene pairs?arrow_forwardLet us suppose that two long-winged flies were crossed and that 77 long-winged and 24 short-winged specimens were counted in the offspring. a. Will the short-winged character be dominant or recessive?B. What will the genotypes of the parents be?C. What is the observed genotype ratio?arrow_forwardA wild-type fruit fly (heterozygous for gray body color andred eyes) is mated with a black fruit fly with purple eyes. Theoffspring are wild-type, 721; black purple, 751; gray purple, 49;black red, 45. What is the recombination frequency betweenthese genes for body color and eye color? Using informationfrom problem 3, what fruit flies (genotypes and phenotypes)would you mate to determine the order of the body color, wingsize, and eye color genes on the chromosome?arrow_forward
- Calculate the recombination frequency from the following data collected from a breeding experiment with butterflies. Parent 1: dihybrid, long legs and blue wings. Parent 2: double recessive, short legs and white wings Offspring: 50 long legs and white wings 200 long legs and blue wings 50 short legs and blue wings 200 short legs and white wings Recombination frequency is: Select one: a. 0.25 (25%) b. 0.10 (10%) c. 0.20 (20%) d. 0.33 (33%) e. 0.50 (50%)arrow_forwardCreate monohybrid crosses from the following information: 33. Homozygous monkey toes with non-monkey toes. 34. What are the genotypes and phenotypes of number 33? 35. Both heterozygous brown eyes. 36. What are the genotypes and phenotypes of number 35? 37. Heterozygous large size with small size. 38. What are the genotypes and phenotypes of number 37?arrow_forwardImagine you are studying a population of clownfish and have genotyped two biallelic loci. The first locus has alleles with frequencies 0.5 and 0.5 in the population. The second locus has alleles with frequencies 0.2 and 0.8. You have genotyped one young fish at these two loci. a. What is the probability that we can exclude a father as the true parent (assuming that he is not the true parent)? b.If you could genotype a third locus to increase your probability of paternal exclusion, would you prefer a locus with allele frequencies of 0.55/0.45 or 0.2/0.8? Why? c.Is the power from part a) good enough to use for a paternity exclusion study without adding other loci? Why or why not?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning