Question 11What statement is FALSE regarding the use linkage mapping for human genetics?O The first linkage map showed the relative location of genetic markers.OA large number of chromosomes means that many traits assort independently, frustrating linkage mapping efforts.O Linkage mapping continues to be the preferred method to determine alleles associated with disease.Initial linkage mapping used family pedigrees.Question 12Which of the following is not a component of Sanger sequencing?

Detailed explanation:a) The first linkage map showed the relative location of genetic markers: This…

Answered: Question 11 What statement is FALSE…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Question 17 How do we go from a genetic association to a causal relation between a gene and a phenotype? ◇ We can use a GWAS with many different samples to identify causal alleles. O We can quantify gene expression of a gene of interest and determine the relationship with a phenotype. O We can determine how often an associated allele occurs in cases that have a disease, which will allow us to determine a causal relationship. By silencing a gene, we can study the effect of that gene on the phenotype. Question 18 What is the function of the CRISPR-Cas system in nature? O part of the bacterial immune response against viral infection a molecule involved in regulation of expression
QUESTION 6 Genes a, b, and c are linked, and pairwise crosses gave the following distances: a-b 20 cM b-c 2 cM a-c 22 cM Which gene is in the middle? O a O It's not possible to tell without additional information. QUESTION 7 A man and woman both have tpe AB blood and are carriers for the Bombay mutation. What is the probability their first child will be phenotypically O (Bombay)? O 1/4 O 1/16 O 1/2 O 3/16 QUESTION 8 The Bombay phenotype occurs because: O Rh antibodies can travel across the placenta and cause an incompatibility reaction. O an individual is homozygous recessive at the Rh locus. O the H substance is not present and so no A or Bantigens can be produced. O an individual does not haveor alleles, and so cannot produce the corresponding antigens.
QUESTION 7 A trihybrid individual with the genotype AaDdEe is testcrossed with an aaddee individual. The resulting offspring are as follows: ADe 880 aDE 67 AdE 45 ade 4 ADE 3 aDe 57 Ade 57 adE 887 Total = 2000 a) Determine the allele arrangement and gene order in the trihybrid parent. Give your answer using the PQr/pqR notation. b) 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)
| Choose ) [ Choose] metagenomics whole-genome shotgun approach bioinformatics ed GenBank genomics
QUESTION 3 Linkage disequilibrium is defined by which of the following? O Positions in the genome where people vary by a single nucleotide base O The occurrence of two alleles in the repulsion configuration O The probability that two genes are linked O Crossing over that occurs between two genes that are located close to each other O The nonrandom association between alleles in a haplotype QUESTION 4 Red-green color blindness is a human recessive X-linked disorder. A young man with a 47, XXY karyotype (Klinefelter syndrome) is colorblind. His mother and father have normal color vision, but his mother has a father who is colorblind Did the nondisjunction event that gave rise to the young man with Klinefelter syndrome happen in the mother, the father or is it impossible to tell (assume that no crossing over took place in prophase I of meiosis)? a. Mother b. Father c. Impossible to tell Did this nondisjunction happen during Meiosis I, Meiosis II, or is it impossible to tell? a. Meiosis I…
QUESTION 8 Check all the statements that are TRUE regarding 16S or ITS (microbiome) illumina (NGS) sequencing vs. whole genome illumina (NGS) sequencing. O A. You don't need to trim the reads for 16S sequencing, but you do for whole genome sequencing. O B. Whole genome sequencing files will have more reads in them because genomes are bigger than 165 amplicons. O C. The read alignment and contig assembly steps would probably be harder for whole genome sequencing. O D. The 165 molecules being sequenced are always DNA, while for whole genome sequencing they could be RNA or DNA going into the illumina sequencer. O E. NGS for 16S and whole genome sequencing can both use sample indexes/barcodes to maximize efficiency. O F. Whole genome sequencing doesn't necessarily require you to know any of the sequences/targets before hand to design specific primers for.
Question 8 Review Figure 21.14 A proposal model for the evolution of globin gene families. Which statement is true about the evolution of the globin gene families? all of these O duplication of ancestral genes O mutation transposition to different chromosome O further duplications and mutations
Question 7 Review non Mendelian genetics. Match the example and its genetics. Each term can only be used once. The coat color of Labrador retrievers and many other mammals is an example of | Choose J The human height and skin color are example of this inheritance. | Choose ABO blood group is an example of | Choose J Snapdragon flower color is an example of [ Choose The hydrangeas flower color growing in different pH environments has a different color. This is an example of | Choose )
Question 9
Question 10 of 18 Attempt 10 Suppose a geneticist uses a three-point testeross to map three recessive, linked hummingbird wing-shape mutations called tk, f, and s, where tk is associated with thick wings, f is associated with flat wings, and s is associated with sharp wings. He first crosses homozygous sharp birds to homozygous thick, flat birds. Next, he testcrosses the Fi progeny to thick, flat, sharp birds. He obtained the results shown. Given this data, label the progeny phenotypic classes as either parental, single crossover (SCO), or double crossover (DCO) recombinant types. Each label may be used multiple times. Two classes have already been filled in. Phenotype Number Recombinant type sharp 819 Answer Bank thick, flat 821 SCO DCO flat 183 parental thick, sharp 169 SCO thick 6. flat, sharp thick, flat, sharp 28 SCO wild type 26
Question 01 Are the SNPS associated with the disease? For each SNP you will have to perform a Chi-Square test. If the SNP is not associated with the disease, there will be a proportional number of affected and unnafected samples for each genotype
Question 1 Being able to associate a QTL with a marker relies on linkage. Suppose researchers genotype a population for two marker loci that they think are on the same chromosome. Using these two markers they notice that out of 1,000 genotypes, there were 53 recombinant individuals. a) Make a conservative estimate of how far apart the two loci are. b) Presuming these are mammals, how many DNA bases does this distance represent?

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