Pearson eText Genetic Analysis: An Integrated Approach -- Instant Access (Pearson+)
3rd Edition
ISBN: 9780135564172
Author: Mark Sanders, John Bowman
Publisher: PEARSON+
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Chapter E, Problem 9P
Additional STR allele frequency information can be added to improve the analysis in Problem
E.8 Figure
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Please discuss the following three
scenarios (A, B and C) and their
product(s)/outcome due to the single base
mutation on Allele 2.
(two sentences max for each scenario,
please)
A
Allele 1
MNMNM
Allele 2
MAMM
Allele 1
B
MNMNM
Allele 2
MAMM
Allele 1
C
MMM
Allele 2
MMM
Hardy-Weinberg Problems
Please be sure to SHOW ALL WORK in order to get credit. This assignment will be graded for accuracy. *Hint: Remember what p and q represent. Determine first if I am asking for allele frequency or genotype/phenotype frequency. Then determine which equation is appropriate.
What does p and q represent in the equations?
Which equation do you use if I am asking for allele frequency?
If the frequency of the recessive allele is 0.1, what is the frequency of the dominant allele?
If the frequency of the dominant allele is 0.4, what is the frequency of the recessive allele?
If the frequency of the dominant allele is 0.6, what is the frequency of the homozygous dominant genotype?
If the frequency of the homozygous recessive genotype is 0.81, what is the frequency of the recessive allele?
If the frequency of the homozygous dominant genotype is 0.25, what is the frequency of the dominant allele?
If the frequency of the…
3. Animation 12.1: This table shows results from allele-specific oligonucleotide hybridization analyses conducted on three different patients. The analyses all used the same probes constructed to test for the presence of a
mutant gene. A plus sign (+) indicates hybridization occurred and a negative sign (-) indicates no hybridization occurred.
Patient #1
Patient #2
Patient #3
Probe for normal
allele
Probe for mutant
allele
Which patient(s) is/are homozygous for the mutation and which patient(s) is/are heterozygous for the mutation?
O Patients #2 and #3 are homozygous and Patient #1 is heterozygous for the mutation.
O Patient #2 is homozygous and Patient #1 is heterozygous for the mutation.
O Patient #1 is homozygous and Patients #2 and #3 are heterozygous for the mutation.
O Patient # 1 is homozygous and Patient #2 is heterozygous for the mutation.
Chapter E Solutions
Pearson eText Genetic Analysis: An Integrated Approach -- Instant Access (Pearson+)
Ch. E - E.1 What is CODIS? Describe the four most...Ch. E - Compare and constast the terms paternity index...Ch. E - What is exclusion principle? How is it used in...Ch. E - What is the statistical principle underlying...Ch. E - Explain the meaning of identity by descent in the...Ch. E - Prob. 6PCh. E - E.7 The results shown are from a DNA test for four...Ch. E - E.8 Figure illustrates the results of an...Ch. E - E.9 Additional STR allele frequency information...Ch. E - Prob. 10P
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- Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown. How many complementation groups are formed by these eight mutants? (Picture attached) A) 2 B) 3 C) 4 D) 5 E) 6arrow_forwardPlease do part Barrow_forwardPlease see the question attached in the image.arrow_forward
- I need some help with this fill-in-the-blank problem. The answer choices are bolded and bracketed. Please see the attached photo to complete. 1. In the D2S441 locus, Sophie's allele [10, 12] is maternal and her allele 10 is paternal from [Sam or Bill only, Sam or Harry only, Bill or Harry only, Sam Bill or Harry (either 3)]. 2. Sophie's allele 13 for D19S433 is [maternal, paternal] and her allele 14 could have come from [Sam and Bill only, Sam and Harry only, Bill and Harry only, Sam Bill or Harry (either 3)]. 3. In the FGA locus, Sophie's allele [21, 22] is maternal and her other allele could have come only from [Sam, Bill, Harry]. 4. Based on all STRs in the 3 panels we studied, it is clear that [Sam, Bill, Harry] is Sophie's father. He has one allele of Sophie's alleles for all STR loci 5. Would these results stand in court as proof paternity: [Yes or No] 6. This type of DNA profiling can also be used to determine maternity. Is there any doubt that Donna is Sophie's biological…arrow_forwardPlease answer all parts along with the reason. I'll definitely give a like. Thank you in advance! 1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab? 1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that 1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny?arrow_forward33arrow_forward
- PLease help, double and triple check your answers, im using this to study, these questions are NOT graded they are PRACTICE problems. PLease help me with all parts of this questionarrow_forwardB What is the genotype ratio? What is the phenotype ratio? 3. What are the expected genotype and phenotype ratios in the following genetic conditions? Use scratch paper to do the Punnett square if needed but you do not need to draw it on the worksheet. a. Monohybrid cross between 2 heterozygous individuals (Aa x Aa) Genotype ratio: 1:2:1 Phenotype ratio: 3:1 G q 8:32 a b. Dihybrid cross between 2 heterozygous individuals (AaBb x AaBb) Genotype ratio: Phenotype ratio: 4. Both Mrs. Smith and Mrs. Jones had babies the same day in the same hospital. Mrs. Smith took home a baby girl, whom she named Sharon. Mrs. Jones took home a girl, whom she named Jane. Mrs. Jones began to suspect, however, that the child had been accidentally switched with Mrs. Smith baby in the nursery. Blood test were made; Mr. Smith was type A, Mrs. Smith was type B, Mr. Jones was type A, Mrs. Jones was type A. Sharon was type O, and Jane was type. B. Had a mix-up occurred? Use scratch paper, you do not need to…arrow_forwardUse the reference to draw question 1 send me a picture of it or make it as text drawing but show it and complete the rest for question 1. Use the information given and your burgeoning knowledge of gene mapping to answer the problems.arrow_forward
- Short hair (S) in rabbits is dominant over long hair (s). The following crosses are carried out, producing the progeny shown. Give all possible genotypes of the parents in each cross.arrow_forwardThe DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●arrow_forwardThe DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non-shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. Given the information above I calculate the level of penetrance seen in image B to be "Blank" 1 percent.arrow_forward
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