In humans, dosage compensation is accomplished by: inactivating one X chromosome in female somatic cells inactivating one homolog from each homologous pair of chromosomes in female somatic cells inactivating the Y chromosome in male somatic cells increasing gene expression from the X chromosome in male somatic cell
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In humans, dosage compensation is accomplished by:
inactivating one X chromosome in female somatic cells
inactivating one homolog from each homologous pair of chromosomes in female somatic cells
inactivating the Y chromosome in male somatic cells
increasing gene expression from the X chromosome in male somatic cell
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- The dominant condition elliptocytosis causes red blood cells to become misshapen into oval-shaped cells. One of the genes responsible for the abnormal shape encodes the band 4.1 protein that together with ankyrin and other scaffold proteins creates and maintains the spherical concave shape of a normal red blood cell. The gene for band 4.1 protein, EPB41, is found on the p arm of chromosome 1. This is very close to the gene encoding the red blood cell Rhesus (Rh) blood type, either phenotype + (dominant) or - (recessive), with a recombination frequency of 2%. This means that 98% of the time alleles for these two genes are linked and are transmitted together. Diane and Jack are siblings, and both have elliptocytosis and Rh+ blood type. Due to the elliptocytosis, both had emergency splenectomies after having severe anemia. Their younger brother, Devonté, has not yet shown signs of elliptocytosis, but has Rh- blood. André, their dad, also has elliptocytosis and Rh+ blood; while their…There are two genetic disorders that result from mutation in imprinted genes: Prader-Willi syndrome, Angelman syndrome. Angelman syndrome results from deletion of UBE3A, which is a gene imprinted such that only the maternal copy is expressed. In the pedigree above, individual I-1 is heterozygous for a deletion of UBE3A and does not have Angelman syndrome. Individual I-2 is homozygous wild type for UBE3A. Which individuals in the pedigree are at risk for exhibiting Angelman syndrome, if any? (Who could potentially have the syndrome, based on what alleles it is possible for them to inherit and express?) Question 8 options: Only I-1 could have been at risk. If he does not have the syndrome, no one in the pedigree could. Only III-1 is at risk I-1, II-2, and III-1 are all at risk Only II-2 is at risk No one in the pedigree is at risk Both II-2 and III-1 are at…Duchenne Muscular Dystrophy (DMD) is a disorder that primarily affects the function of skeletal muscles used for movement and cardiac muscles used for heart beating. Dystrophin is a protein encoded by a single gene, DMD, that is expressed in skeletal and cardiac muscle. Some forms of muscular dystrophy may be caused by different mutations in the DNA sequence of the DMD gene. Because the DMD locus is on the X chromosome, males are affected at higher rates. Two brothers, one of whom has DMD and one of whom does not, worked with their genetic counselor (Links to an external site.) to have their DMD gene sequenced to identify genetic variation that may explain why one brother was affected and the other not. Because DMD is a very long gene, a fictionalized, simplified model of the results is presented here (Figure 1). The actual DMD mRNA is about 16,000 base-pairs!------Consider single nucleotide polymorphism (SNP) #1 (Figure 1). Is this mutation likely to cause Duchenne muscular…
- 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 inactivationWhen the organization of adjacent genes with highly similar sequences is in an inverted orientation, this can reduce the expression of other genes that have similar sequence and are located on other chromosomes. Explain the mechanism of how this generally occurs.Imagine a scenario in which prenatal testing of a human female fetus indicates that the baby will have a normal XX karyotype but is heterozygous for a mutation that inactivates the Xist promoter. Allele “Xr” represents the mutated version of the Xist promoter, and “XR” represents the normal version of the Xist promoter. How will this mutation affect the process of X inactivation?A. "X inactivation will still be random with both mutant and normal X chromosomes being randomly inactivated" B. "The chromosome with the mutant Xist promoter will always be active" C. "The chromosome with the mutant Xist promoter will always be inactivated" D. "The normal X chromosome (no mutation in Xist) will always be active" E. "The normal X chromosome (no mutation in Xist) will always be inactivated"
- In the Fast Forward Box Visualizing X Chromosome Inactivation in Transgenic Mice, suppose the investigators had looked at the expression of green and red fluorescent protein in early mouse embryos, when the embryos have fewer than 500 cells. What patterns would they likely have observed?Which of the following does NOT pertain to the myoblast-determining gene 1?*a. It is a master gene.b. It is a silencing gene.c. It produces a transactivating protein.d. It activates its own gene. Gene silencing involves which type of histone modification?* a. acetylation of histone 4 b. dimethylation of histone 3 c. trimethylation of histone 4 d. trimethylation of histone 3 Given the required environment, the totipotency of the nucleus can allow which of the following?* a. a committed cell to undergo dedifferentiation b. a committed cell to undergo terminal differentiation c. a terminally differentiated cell to produce a complete organism d. a terminally differentiated cell to produce specific types of tissues An induced pluripotent cell is described by which of the following?* a. It is a committed cell that undergoes redifferentiation. b. It is a committed cell that undergoes dedifferentiation. c. It is a terminally…Which of the following statements is false? Dosage compensation is accomplished in humans by stimulation of gene expression from the single X chromosome. Dosage compensation is accomplished in humans by inactivating a female X chomosome. An individual with Turner Syndrome has no Barr bodies. An individual with Klinefelter syndrome generally has one Barr body. A typical XX human female has one Barr body.
- Duchenne muscular dystrophy (DMD) is an X-linked recessive genetic disease caused by mutations in the gene that encodes dystrophin, a large protein that plays an important role in the development of normal muscle fibers. The Dystrophin gene is immense, spanning 2.5 million base pairs, and includes 79 exons and 78 introns. Many of the mutations that cause DMD produce premature stop codons, which bring protein synthesis to a halt, resulting in a greatly shortened and nonfunctional form of dystrophin. Some geneticists have proposed treating DMD patients by introducing small RNA molecules that cause the spliceosome to skip the exon containing the stop codon (A. Goyenvalle et al., 2004. Science 306:1796–1799). The introduction of the small RNAs will produce a protein that is somewhat shortened because an exon is skipped and some amino acids are missing, but it may still result in a protein that has some function. The small RNAs, antisense RNAs, used for exon skipping are complementary to…Duchenne muscular dystrophy (DMD) is an X-linked recessive genetic disease caused by mutations in the gene that encodes dystrophin, a large protein that plays an important role in the development of normal muscle fibers. The dystrophin gene is immense, spanning 2.5 million base pairs, and includes 79 exons and 78 introns. Many of the mutations that cause DMD produce premature stop codons, which bring protein synthesis to a halt, resulting in a greatly shortened and nonfunctionalform of dystrophin. Some geneticists have proposed treating DMD patients by causing the spliceosome to skip the exon containing the stop codon. Exon skipping would produce a protein that is somewhat shortened (because an exon is skipped and some amino acids are missing), but might still result in a protein that had some function (A. Goyenvalle et al. 2004. Science 306:1796–1799). Propose a possible mechanism to bring about exon skipping for the treatment of DMD.There are two genetic disorders that result from mutation in imprinted genes: Prader-Willi syndrome and Angelman syndrome. Prader-Willi syndrome results from deletion of region 15q11-q13, which in healthy individuals is a region imprinted such that only the paternal copy is expressed. In the pedigree above, individual I-1 is heterozygous for a deletion of region 15q11-q13 and does not have Prader-Willi syndrome. Individuals I-2 and II-1 are both homozygous wild type for the region. Which individuals in the pedigree might have Prader-Willi syndrome? (Who could potentially have the syndrome, based on what alleles it is possible for them to inherit and express?) Question 9 options: Only II-2 could have Prader-Willi syndrome III-1 could have Prader-Willi syndrome in the presented pedigree; II-2 could only have had it if she were male Both II-2 and III-1 could have Prader-Willi syndrome II-2 could have…