Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 11, Problem 29P
One of the difficulties faced by human geneticists is that matings are not performed with a scientific goal in mind, so pedigrees may not always provide desired information. As an example, consider the following matings (W, X, Y, and Z):
| a. | Which of these matings are informative and which noninformative for testing the linkage between anonymous loci A and B? (A1 and A2 are different alleles of locus A, B1 and B2 are different alleles of locus B, etc.) Explain your answer for each mating. |
| b. | Is locus A more l b. ikely to be a SNP or an SSR? What about locus B? Explain. |
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A series of three-point testcrosses is made to determine the genetic map order of seven linked allele pairs: A/a, B/b, G/g, H/h, Q/q, R/r, and Y/y.From each cross between a triply heterozygous parent listed below, two recombinant classes were noticed as the least frequent among all 8 progeny classes, and are listed at the right in the table.
A. For each testcross write the genotype of the F1 heterozygous parent.
F1 Parental Phenotype Least frequent F2 Phenotype
1.AHB&ahb AHb & ahB
2.RYh&ryH RYH & ryh
3.BhY&bHy Bhy & bHY
4.qYB&Qyb qYb & QyB
5.AbQ&aBq Abq & aBQ
6.ghR&GHr ghr & GHR
B. Write the unified map order of these genes, showing your reasoning.
You are studying four linked genes located on chromosome 2 in the fruit fly Drosophila
melanogaster: adp (which contributes to obesity), b (which contributes to body color), pr (which
contributes to metabolism), and vg (which contributes to wing formation). A series of crosses
between pair-wise combinations of these mutations yielded the following recombination frequencies
between the indicated loci:
pr and adp 9%
adp and b 6%
pr and b 3%
vg and b 10%
vg and pr 7%
What is the genetic distance in map units (cM) between the adp and vg loci?
16
9
6
4
A researcher crosses mice with brown eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes:
F1: 6 apricot, short : 30 brown, long : 15 brown, short : 9 apricot, long
You know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these
results.
a) Making use of the appropriate genetic convention for naming alleles, give the genotype of the male parent in this cross.
b) What is your null hypothesis for the chi-square test?
c) Give the expected number of individuals in the "brown, long" class.
d) You obtain a value of 3.47 for the chi-square test. What conclusion can you make from the results of the chi-square test?
P
df
0.995
0.975
0.9
0.5
0.1
0.05*
0.025
0.01
0.005
1
0.000
0.000
0.016
0.455
2.706
3.841
5.024
6.635
7.879
2
0.010
0.051
0.211
1.386
4.605
5.991
7.378
9.210
10.597
0.072
0.216
0.584
2.366
6.251
7.815
9.348
11.345
12.838
4
0.207
0.484
1.064
3.357…
Chapter 11 Solutions
Genetics: From Genes to Genomes
Ch. 11 - Choose the phrase from the right column that best...Ch. 11 - Would you characterize the pattern of inheritance...Ch. 11 - Would you be more likely to find single nucleotide...Ch. 11 - A recent estimate of the rate of base...Ch. 11 - If you examine Fig. 11.5 closely, you will note...Ch. 11 - Approximately 50 million SNPs have thus far been...Ch. 11 - Mutations at simple sequence repeat SSR loci occur...Ch. 11 - Humans and gorillas last shared a common ancestor...Ch. 11 - In 2015, an international team of scientists...Ch. 11 - Using PCR, you want to amplify an approximately 1...
Ch. 11 - Prob. 11PCh. 11 - The previous problem raises several interesting...Ch. 11 - You want to make a recombinant DNA in which a PCR...Ch. 11 - You sequence a PCR product amplified from a...Ch. 11 - Prob. 15PCh. 11 - The trinucleotide repeat region of the Huntington...Ch. 11 - Sperm samples were taken from two men just...Ch. 11 - Prob. 18PCh. 11 - a. It is possible to perform DNA fingerprinting...Ch. 11 - On July 17, 1918, Tsar Nicholas II; his wife the...Ch. 11 - The figure that follows shows DNA fingerprint...Ch. 11 - Microarrays were used to determine the genotypes...Ch. 11 - A partial sequence of the wild-type HbA allele is...Ch. 11 - a. In Fig. 11.17b, PCR is performed to amplify...Ch. 11 - The following figure shows a partial microarray...Ch. 11 - Scientists were surprised to discover recently...Ch. 11 - The microarray shown in Problem 25 analyzes...Ch. 11 - The figure that follows shows the pedigree of a...Ch. 11 - One of the difficulties faced by human geneticists...Ch. 11 - Now consider a mating between consanguineous...Ch. 11 - The pedigree shown in Fig. 11.22 was crucial to...Ch. 11 - You have identified a SNP marker that in one large...Ch. 11 - The pedigrees indicated here were obtained with...Ch. 11 - Approximately 3 of the population carries a mutant...Ch. 11 - The drug ivacaftor has recently been developed to...Ch. 11 - In the high-throughput DNA sequencing protocol...Ch. 11 - A researcher sequences the whole exome of a...Ch. 11 - As explained in the text, the cause of many...Ch. 11 - Figure 11.26 portrayed the analysis of Miller...Ch. 11 - A research paper published in the summer of 2012...Ch. 11 - Table 11.2 and Fig. 11.27 together portray the...Ch. 11 - The human RefSeq of the entire first exon of a...Ch. 11 - Mutations in the HPRT1 gene in humans result in at...Ch. 11 - Prob. 44P
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
- Following the analysis of a pedigree, a genetic link at 4cM is considered between a mutation leading to a pathology and the molecular marker HUMTH01. The study counts 14 "parental" and 3 "recombinant" individuals. We call p(theta=0.04) is the probability of obtaining such a pedigree in case of a 4cM genetic linkage, p(theta=0.5) is the probability of obtaining such a pedigree in case of independence between the mutation and the marker, Z(theta=0.04) the value of the Lod-score under the assumption of 4cM genetic linkage. Tick all the correct answers: p(theta=0,04)=1,79.10E-9 and Z(theta=0,04)=0,47 p(theta=0,5)=7,18.10E-12 and Z(theta=0,04)=0,77 and Z(theta=0,04)=0,67 p(theta=0,04)=2,75.10E-10 p(theta=0,5)=6,04.10E-10 and Z(theta=0,04)=0,47 p(theta=0,5)=9,36.10E-12 and Z(theta=0,04)=1,33 p(theta =0,5)=5,82.10E-11 and Z(theta=0,04)=0,67 no correct answer p(theta=0,04)=4,31.10E-11 p(theta=0,04)=2,01.10E-10 and Z(theta=0,04)=0,77 and Z(theta=0,04)=1,33arrow_forwardConsider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PCR was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PCR products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forwardWhile studying the genetics of the "Cute Faced Bat" face structure gene, three alleles ) are identified that produce 6 genotypes and 4 phenotypes “Cute”, “Super Cute Nose”, “Super Cute Face”, and “Super, Duper Cute” (equally expressing both Super Cute Face and Super Cute Nose). Given these observations, what can you conclude about the allele interactions?arrow_forward
- A trihybrid individual with the genotype AaDdEe is testcrossed with an aaddee individual. The resulting offspring are as follows: ADe 7 aDE 157 AdE 688 ade 153 ADE 125 aDe 680 Ade 182 adE 8 Total = 2000 a) b) Determine the allele arrangement and gene order in the trihybrid parent. Give your answer using the PQr/pqR notation. Draw a chromosome map for these genes. Give your map distances to two decimal places. (Use underline or underscore (_) to draw the line for the map)arrow_forwardSuppose that the wild type phenotype for tomatoes is red, round fruit with three-parted leaves and that a strain of mutant tomatoes has yellow, pear-shaped fruit with entire leaves. A and a represent the alleles for fruit color, B and b represent the alleles for fruit shape, and C and c the leaf shape. Two tomatoes with the genotypes Aa Bb cc and Aa Bb CC are crossed. The resulting Punnett square is shown. One cell has been completed with the correct tomato phenotype. Fill in the rest of the Punnett square by placing a fruit and leaf phenotype in each cell. ABC AbC aBC abC ABC Abc aBc abc Answer Bankarrow_forwardA presumed dihybrid in Drosophila, BbFf is testcrossed with bbff. (B-black body, b-brown body, F-forked bristles, f-unforked bristles) The results for the testcross are below: Black, forked: 230 Black, unforked: 210 Brown, forked: 240 Brown, unforked: 250 Determine the expected values and ratios for each genotype.arrow_forward
- A researcher crosses mice with brown eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes:F1: 6 apricot, short : 30 brown, long : 15 brown, short : 9 apricot, longYou know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these results.a) Making use of the appropriate genetic convention for naming alleles, give the genotype of the male parent in this cross. b) What is your null hypothesis for the chi-square test? c) Give the expected number of individuals in the “brown, long” class. d) You obtain a value of 3.47 for the chi-square test. What conclusion can you make from the results of the chi-square test?arrow_forwardKernel color in wheat is controlled by 2 pairs of genes (AABB). Determine the color of the offspring with the following genotypes: (Note: 4 contributing alleles – red; 3 contributing alleles – medium red; 2 contributing alleles – intermediate red; 1 contributing allele – red; and 0 – white (AAbb x AaBb, AABb x Aabb, aaBb x Aabb, AABb x aabb, AABb x AaBb) Show the solution through punnett square. Refer to the problem above, a wheat plant producing medium red seed is crossed with another plant producing intermediate red seed. How many individuals will be?Show the solution through punnett square. Red Medium red Intermediate red Light red Whitearrow_forwardA female from true breeding line of Drosophila with white eyes is crossed with a male from a true breeding line with brown eyes. All of the offspring have wild type brick red eyes. Which of the following explanations is most likely? A) There are many alleles for the single gene for eye color. Wild type brick red eyes result only when the fly is heterozygous. B) The alleles for brown, white, and brick red eyes are alleles for a single locus. The allele for brown eye color is dominant to the allele for brick red eye color and to the allele for white eyes. C) There is more than one gene for eye color. The brown mutation and the white mutation occur in separate genes and are both recessive to the wild type alleles. The offspring are heterozygous for both genes, so they are phenotypically wild type. D) None of the above. It is not possible for a cross between a white-eyed and a brown-eyed fly to produce wild type offspring.arrow_forward
- Kernel color in wheat is controlled by 2 pairs of genes (AABB). Determine the color of the offspring with the following genotypes: (Note: 4 contributing alleles – red; 3 contributing alleles – medium red; 2 contributing alleles – intermediate red; 1 contributing allele – red; and 0 – white (AAbb x AaBb, AABb x Aabb, aaBb x Aabb, AABb x aabb, AABb x AaBb) Refer to the problem above, a wheat plant producing medium red seed is crossed with another plant producing intermediate red seed. How many individuals will be? Red Medium red Intermediate red Light red Whitearrow_forwardYou are mapping three linked loci in Drosophila melanogaster (the common laboratory fruit fly). You cross flies that are triply mutant for apricot (pale eyes), bristle (extra bristles) and clipped (notched wings) to wild-type flies. The F+ flies are wild-type in appearance. You then backcross the F+ females to pure-breeding (apricot, bristle, clipped) males and score the phenotypes of 1000 F progeny for all three loci. Here are the results: 359 wild-type 361 apricot, bristle, clipped 89 bristle, clipped 91 apricot 42 apricot, bristle 38 clipped 9 apricot, clipped 11 bristle Using these data, first determine what gametes from the F; trihybrid produced each of the eight F2 categories. Note that apricot = aa (recessive to wild-type A); bristle = bb (recessive to wild-type B); and clipped = cc (recessive to wild-type C). Then determine if each gamete is recombinant (R) or nonrecombinant (R) for each pair of alleles (that is, for each genetic interval). Complete the table by dragging the…arrow_forwardIn a three point cross problem for genes Q, L, and T, you have identified the parentals and double recombinants. Using that data (shown below), what is the order of the genes? Type: Alleles parental q L t parental Q l T double cross over q L T double cross over Q l t Group of answer choices L Q T L T Q T L Q the correct answer is not availablearrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSON
Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Anatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education
Human Anatomy & Physiology (11th Edition)
Biology
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:PEARSON
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Anatomy & Physiology
Biology
ISBN:9781259398629
Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa Stouter
Publisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)
Biology
ISBN:9780815344322
Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter
Publisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & Physiology
Biology
ISBN:9781260159363
Author:Martin, Terry R., Prentice-craver, Cynthia
Publisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)
Biology
ISBN:9781260231700
Author:Sylvia S. Mader, Michael Windelspecht
Publisher:McGraw Hill Education
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