Question 17How 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 18What is the function of the CRISPR-Cas system in nature?O part of the bacterial immune response against viral infectiona molecule involved in regulation of expression

The correct answer is (d) By silencing a gene, we can study the effect of that gene on the…

Answered: Question 17 How do we go from a genetic…

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 11 What 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 12 Which of the following is not a component of Sanger sequencing?
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- QUESTION 1 Consider three linked genes: A, B, C. Percent recombination between gene B and C is 11% and no crossing over occurs between A and B. If a heterozygote in cis conformation is test crossed, what is the probability of an offspring with the genotype AaBbcc? A. 0.011 B. 0.055 C.0.022 D.0.089 E. 0.045
Question GROUP A: Genetics Problems In humans, red-green color blindness is caused by a recessive allele on the X chromosome. A male with normal vision and his color-blind wife have a child. If the child is male what is the probability that he will be color-blind? The female in the previous problem has a daughter that is color-blind. The husband claims that this cannot be his child. Can you support or refute his argument?
Quèstion 20 Selection coefficient measures the selective disadvantage of a disfavored genotype. Functional Genomics elucidates the patterns of coordinate gene regulation. The 1st statement is True and the 2nd statement is False. The 1st statement is False and the 2nd statement is True. O Both Staments are True. O Both statements are False.
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 9 Which of the following statements about Genome-wide association studies is NOT true? O A genome-wide association study is an approach that involves testing genetic variations in the genomes of many people to find genetic variations associated with a particular disease. O When DNA variations are linked to a trait through a genome-wide association study, they always exhibit an autosomal recessive inheritance pattern. O Genome-wide association studies are useful because they can identify alleles that have small effects on multifactorial traits. O Genome-wide association studies only became possible in the past 5-10 years when methods such as SNP-chip and whole genome sequencing were invented.
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 #1 a. Explain why there is a non-Mendelian 2:1 yellow/agouti phenotypic ratio in this cross. b. What is the Raly gene? What does it encode? How does it relate to the lethality of Ay? c. A heterozygous mouse AAy does not produce enough pigment and it is yellow. Homozygous individuals die during development, but AAy heterozygous do not. Why? You may use any graphic aid to explain this, if necessary. This means that you may use a Punnett square or a forking segregation diagram, or computations with fractions.
QUESTION 2 A researcher crosses mice with brown eyes and short tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 32 apricot, short : 30 brown, long : 139 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) Give the expected number of individuals in the "brown, short" class. b) What is your null hypothesis for the chi-square test? c) You obtain a value of 8.59 for the chi-square test. What conclusion can you make from the results of the chi-square test? d) Making use of the appropriate genetic convention for naming alleles, give the genotype of any of the progeny in the "apricot, short" class. 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…
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QUESTION 3 You performed a Genome-Wide Association Study (GWAS) attempting to link genotypes to the likelihood of having Disease X. You found that the Odds Ratio for those with Genotype A relative to Genotype B for having Disease X to be equal to 3. How would you best describe the risk of having Disease X with Genotype A compared to Genotype B? A Equal risk for Disease X with Genotype A B. Less risk for Disease X with Genotype A OC. More risk for Disease X with Genotype A OD. No way to determine which genotype has more risk

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