afa Q a/a Q Unsupplemented Supplemented b A"/a offspring Unsupplemented mother Supplemented mother This question relates to the aqouti mice research discussed in the lecture. What did the researchers observe in the experimental group (AVYAX aa) that was fe methyl donors? They gained weight rapidly. Increased expression of aqouti gene. Reduced risk of chronic disease Higher percentage of yellow offspring. ViV

afa Q a/a Q Unsupplemented Supplemented b A"/a offspring Unsupplemented mother Supplemented mother This question relates to the aqouti mice research discussed in the lecture. What did the researchers observe in the experimental group (AVYAX aa) that was fe methyl donors? They gained weight rapidly. Increased expression of aqouti gene. Reduced risk of chronic disease Higher percentage of yellow offspring. ViV

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 20QP: The cystic fibrosis gene encodes a chloride channel protein necessary for normal cellular functions....

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Glucose-6-phosphate dehydrogenase deficiency/G6PDD (g) is an X-linked recessive conditionwherein the red blood cells of affected individuals undergo premature hemolysis. Fragile X syndrome(F), on the other hand, is an X-linked dominant mutation characterized by a mild to moderateintellectual disability. Amelogenesis imperfecta (A) is a congenital disorder affecting the formation ofthe teeth enamel making affected individuals at higher risk for dental cavities and related problems.Only male offspring inherit this condition. Jane is heterozygous for both X-linked traits like her mother. Her father is normal for both X-linked traits. James has a mother who suffers from G6PDD but not from fragile X-syndrome. His fatherdoes not exhibit any X-linked disorder but has amelogenesis imperfecta. A. What are the genotypes of the following: Jane: ______________________ James: ______________________Jane’s mother: _________________ James’ mother: ________________Jane’s father: __________________…
a. The ABO blood groups are controlled by the alleles of a single gene. A very rare example of epistasis is called the Bombay phenotype. If many couples with the genotypes IAIBHh x IAiHh have children, what ABO phenotypic ratio would you expect in their offspring? b. What genetic term (excluding epistasis) can be used to describe the expression of the ABO alleles in the presence of the hh genotype?
. Neurofibromas are tumors of the skin that can arisewhen a skin cell that is originally NF1+/ NF1− losesthe NF1+ allele. This wild-type allele encodes a functional protein (called a tumor suppressor), while theNF1− allele encodes a nonfunctional protein.A patient of genotype NF1+ / NF1− has 20 independent tumors in different areas of the skin. Samplesare taken of normal, noncancerous cells from thispatient, as well as of cells from each of the 20 tumors.Extracts of these samples are analyzed by a techniquecalled gel electrophoresis that can detect variantforms of four different proteins (A, B, C, and D) allencoded by genes that lie on the same autosome asNF1. Each protein has a slow (S) and a fast (F) formthat are encoded by different alleles (for example, ASand AF). In the extract of normal tissue, slow and fastvariants of all four proteins are found. In the extractsof the tumors, 12 had only the fast variants of proteinsA and D but both the fast and slow variants of proteins B and…
The expression of antigen A or antigen B in red blood cells requires the help of an H antigen. A recessive mutation (h) that prevents the synthesis of the H antigen also prevents the expression of A and B antigens. This is called the Bombay effect. There is no ill effect in an individual with this mutation, but complications with blood transfusions or parental disputes may arise. b. Eva’s parents, John and Cher. Are both type AB. John questioned Cher’s fidelity when Eva turnedout to be a type O. Assuming that Cher is indeed loyal to her husband, why did Eva turn out to betype O? Help Cher prove her innocence by showing the possible genotypes of John and Cher. Show alsothe COMPLETE cross that produced Eva.
The expression of antigen A or antigen B in red blood cells requires the help of an H antigen. A recessive mutation (h) that prevents the synthesis of the H antigen also prevents the expression of A and B antigens. This is called the Bombay effect. There is no ill effect in an individual with this mutation, but complications with blood transfusions or parental disputes may arise. a. Individuals with the Bombay genotype (hh) produce anti-H antigen. How can this be a problem during blood transfusion?
The three homeodomain proteins Abd-B, Abd-A, andUbx are encoded by genes within the Bithorax complexof Drosophila. In wild-type embryos, the Abd-B gene isexpressed in the posterior abdominal segments, Abd-Ain the middle abdominal segments, and Ubx in the anterior abdominal and posterior thoracic segments. Whenthe Abd-B gene is deleted, Abd-A is expressed in both themiddle and the posterior abdominal segments. WhenAbd-A is deleted, Ubx is expressed in the posterior thoraxand in the anterior and middle abdominal segments.When Ubx is deleted, the patterns of Abd-A and Abd-Bexpression are unchanged from wild type. When bothAbd-A and Abd-B are deleted, Ubx is expressed in all segments from the posterior thorax to the posterior end ofthe embryo. Explain these observations, taking into consideration the fact that the gap genes control the initialexpression patterns of the homeotic genes
The ABO blood groups in humans are expressed as the IA,IB and i alleles. The IAallele encodes the A blood group antigen, IB encodes B, and i encodes O. Both A and B are dominant to 0. If a heterozygous blood type A parent (IAi) and a heterozygous blood type B parent (IBi) mate, one quarter of their offspring are expected to have the AB blood type IAIBin which both arnigens are expressed equally. Therefore, ABO blood groups are an example of: a. multiple alleles and incomplete dominance b. codominance and incomplete dominance c. incomplete dominance only d. multiple alleles and codominance
Two genes associated with breast cancer, BRCA1 and BRCA2, were discovered in 1994 and 1995, respectively, and shortly thereafter, were patented by Myriad Genetics, a company based in Utah. Under the patents, testing for mutations in these genes could only be performed by Myriad, at costs from 300 to 3,000. Myriad also patented the process of analyzing the results of such tests, preventing anyone who obtains the sequence of their BRCA genes by other means (which itself would probably be patent infringement) from interpreting the information. The idea that genes can be patented has been a contentious issue from the beginning. Patents are not granted for products of nature, meaning that genes inside the body are not patentable, but biotech companies successfully argued that by removing a gene from the human body, purifying it, and then obtaining its DNA sequence, they created something not found in nature, and which is therefore a patentable invention. The U.S. Patent Office found the argument persuasive, but opponents argue that genes are parts of our bodies and can be identified but not invented. Biotech companies argue that without the protection offered by patents, they would have no incentive for research and development of diagnostic tests. In Europe, patents for BRCA1 and BRCA2 were revoked in 2004 because they did not meet the standards for a patent. After more than a decade of legal disputes, the patents were partially restored in 2008 on a very restricted basis. In the United States, a lawsuit, focused on the patents for the BRCA genes, was filed in May 2009. The suit challenges the basic idea that genes are patentable. In November 2009, the judge ruled that the lawsuit can proceed, and the case is moving forward. In March 2010, a federal court invalidated Myriad Genetics patent on these genes. In August 2011, the U.S. Court of Appeals reversed the lower courts decision and ruled that gene sequences isolated from cells are not a product of nature and are therefore patentable. The case went to the U.S. Supreme Court, which ordered the appeals court to reconsider the case. The Federal Appeals Court did not change its decision, and the case once again, went to the U.S. Supreme Court. A unanimous decision in June 2013 invalidated Myriads patents on the basis that isolating a gene from nature does not make it patentable. This is a landmark decision on gene patenting with widespread ramifications for the biotechnoloogy industry. Will this decision reduce the incentives for companies to invest in new diagnostic tests that would be used by cancer victims or those with serious genetic disorders?
The cystic fibrosis gene encodes a chloride channel protein necessary for normal cellular functions. Let us assume that if at least 5% normal channels are present, the affected individual has mild symptoms of cystic fibrosis. Having less than 5% normal channels produces severe symptoms. At least 50% of the channels must be expressed for the individual to be phenotypically normal. This gene has various mutant recessive alleles: Predict the percent of functional channels and severity of symptoms for the following genotypes: a. heterozygous for CF100 b. homozygous for CF100 c. heterozygous, with one copy of CF100 and one of CF3 d. heterozygous, with one copy of CF1 and one copy of CF3
Are the following events best explained by mutation or epimutation? A. imprinting of the Igf2 geneB. variation in coat color in mice carrying the Avy alleleC. formation of cancer cellsD. variation in flower color between different strains of pea plants,such as purple versus whiteE. X-chromosome inactivation
Two male mice, which we will call male A and male B, are bothphenotypically normal. Male A was from a litter that containedhalf phenotypically normal mice and half dwarf mice. The motherof male A was known to be homozygous for the normal Igf2 allele.Male B was from a litter of eight mice that were all phenotypicallynormal. The parents of male B were a phenotypically normal maleand a dwarf female. Male A and male B were put into a cage withtwo female mice that we will call female A and female B. FemaleA is dwarf, and female B is phenotypically normal. The parents ofthese two females were unknown, although it was known that theywere from the same litter. The mice were allowed to mate witheach other, and the following data were obtained:Female A gave birth to three dwarf babies and four normal babies.Female B gave birth to four normal babies and two dwarf babies.Which male(s) mated with female A and female B? Explain.
Create a typewritten document providing answers to these questions Questions: 1. How does the genetic code determine the expression of heritable traits in an organism? 2. What are the mechanisms of gene regulation that control the expression of heritable traits? 3. What are the functions of DNA segments that do not code for proteins?