In the accompanying diagram, the cate the positions of restriction sites in two alterna- tive DNA fragments that can be present at a locus in a human chromosome. An RFLP analysis is carried out, using probe DNA that hybridizes with the frag- ments at the position shown by the rectangle. With respect to this RFLP, how many genotypes are pos- sible? (Use the symbol A, to refer to the allele that yields the upper DNA fragment, and A, to refer to

In the accompanying diagram, the cate the positions of restriction sites in two alterna- tive DNA fragments that can be present at a locus in a human chromosome. An RFLP analysis is carried out, using probe DNA that hybridizes with the frag- ments at the position shown by the rectangle. With respect to this RFLP, how many genotypes are pos- sible? (Use the symbol A, to refer to the allele that yields the upper DNA fragment, and A, to refer to

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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Homologous recombination in E. coli forms heteroduplex regions of DNA containing mismatched bases. Why are these mismatches not eliminated by the mismatch repair system?
You are a researcher studying a gene you think is responsible for super human strength. You call the gene HULK1. Below is the ORF of this gene from a normal strength individual (with nucleotide position indicated above). Use this sequence to answer the following questions. 1 10 20 30 39 5'- ATG TCA AAG GAC ATT CGC TGT GGG CTC AGA CTA TCA TAA -3' You design primers (6nt each) to amplify this entire sequence. Which primer set below did you use? O a. 5-TTATGA-3' and 5'-TGACAT-3' Ob.5'-TGACAT-3 and 5'-TCATAA-3' Oc 5'-ATGTCA-3' and 5'-TCATAA-3' Od. 5'-ATGTCA-3' and 5'-TTATGA-3' Oe. 5'-ATGTCA-3' and 5'-TGACAT-3'
1) Using the diagram below, sketch in the pattern of bands you would expect to see after digesting the DNA of the TT, Tt and tt genotypes of the TAS2R38 gene. Use the Base Pair Standards on the left of the diagram as a reference in drawing the positions of the bands. Base Pair TASTER GENO TYPE Standards (Base Pairs) TT Tt tt 500 400 300 200 100 50
The region of the normal hemoglobin gene used for genetic testing for sickle cell anemia contains a restriction site such that homozygous normal individuals show two DNA fragments. If a single nucleotide change in hemoglobin destroys that restriction site, then how many DNA fragments will be visible on a gel from individuals that are homozygous mutant? What about heterozygotes?
Three mutations were obtained in a bacterial gene. An antibody is available for the protein product of this gene. Both Northern analysis (RNA separated by electrophoresis, blotted, and probed with DNA) and Western analysis (proteins separated by electrophoresis, blotted, and probed with antibodies) were performed on the mutants. The results are summarized below. Northern Size Western Size 1 2 3 1 2 3 + Long Short Short Long For each mutation, what kind of mutation occurred and how do you know? a) Mutant 1 b) Mutant 2 c) Mutant 3
Nucleosomes can be assembled onto defined DNA segments. When a particular 225-bp segment of human DNA was used to assemble nucleosomes and then incubated with micrococcal nuclease, which digests DNA that is not located within the nucleosome, uniform fragments 147 bp in length were generated. Subsequent digestion of these fragments with a restriction enzyme that cuts once within the original 225-bp sequence produced two well-defined bands at 37 bp and 110 bp. Why do you suppose two well-defined fragments were generated by restriction digestion, rather than a range of fragments of different sizes? How would you interpret this result?
It is possible to "reconstitute" nucleosomes by mixing DNA and his- tone octamers in 2M NaCl and then dialyzing to low salt. When such experiments were carried out using a specific 208-bp fragment of sea urchin DNA, the following results were obtained: Digestion of the product with micrococcal nuclease gave quantitative production of 146-bp DNA. Upon cleavage of this DNA with a restriction nucle- ase having a single site in the fragment, several sharp bands were obtained, with sizes as follows: 29 bp, 39 bp, 107 bp, 117 bp. How would you interpret these data in terms of nucleosome positioning on this DNA?
Calculate the number of Alu elements present in the haploid human genome based on the number of Alu elements present in the clones analysed by the entire class. You will need to know the total length of DNA analysed and the number of Alu elements present. For the entire class, 57 Alu elements are present in 174,204 bp.
As the leading scientist in a biomedical science laboratory, it is a requirement to give advice to your lab assistants when they are having problems with their experiments. What advice would you give to your assistants that are having the following problems: After performing a polymerase chain reaction (PCR) and agarose gel electrophoresis to confirm the presence of the C01 gene of 750bp. 2.1. They observe no band appearing on an agarose gel. What would be your conclusion? 2.2. They observe three bands of different sizes that resemble a smear on the gel. Advice 2.3. They observe a single band on the gel and conclude that the PCR product is an exact copy of the original template DNA. Would you support their condusion? Explain. 2.4. Explain how PCR can be used to detect infectious agents in diagnoses of diseases.
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A pair of paralogous repeats, A and B, have 96% sequence similarity and therefore can promote non-allelic homologous recombination (NAHR). They exist in four possible arrangements in a genome, illustrated below as arrangements 1 – 4. What is the result of NAHR between repeats A and B in arrangement 1? A.Translocation between chromosomes 1 and 2 resulting in monocentric chromosomes B.Deletion or duplication of the region between A and B C.Translocation between chromosomes 1 and 2 resulting in acentric and dicentric chromosomes D.Inversion of the region between A and B