Concept explainers
To analyze:
Nine base pair deletion in the genome of mitochondria between the cytochrome oxidase subunit II gene and
Introduction:
The examination of variations in genome of mitochondria in human populations gives us an idea about the journey of human ancestors throughout the world (out of Africa).The variation pattern of mitochondrial DNA indicates the African origin and journey of modern humans throughout the world.
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Genetic Analysis: An Integrated Approach (3rd Edition)
- If you compare the frequency of the sixteen possible dinucleotide sequences in the E. coli and human genomes, there are no striking differences except for one dinucleotide, 5ʹ-CG-3ʹ. The frequency of CG dinucleotides in the human genome is significantly lower than in E. coli and significantly lower than expected by chance. Why do you suppose that CG dinucleotides are underrepresented in the human genome? (hint: The C in the CG pair is often methylated). Explain how this observation has an impact on the cells immune response.arrow_forward54. 35delG mutation in the connexin 26 gene is an autosomal recessive condition that has been identified to cause congenital deafness. In a population of 367 Greek children, 75 of the children carried two copies of the 35delG mutation.In order to calculate the percentage of individuals who are heterozygous for the 35delG mutation, the calculation that must be performed is Select one: a. 1 – (p2 – 2pq) b. 1 – (p +q) c. 1 – (p2 + q2) d. 1 – q2arrow_forwardThe human genome contains approximately 106 copies of an Alusequence, one of the best-studied classes of short interspersedelements (SINEs), per haploid genome. Individual Alus share a282-nucleotide consensus sequence followed by a 3'-adeninerichtail region. Given that there are approximately 3 * 109bp per human haploid genome, about how many base pairs arespaced between each Alu sequence?arrow_forward
- Mitochondrial DNA sequences have been detected in the nuclear genomes of many organisms, and cpDNA sequences are sometimes found in the mitochondrial genome. Propose a mechanism for how such “promiscuous DNA” might move between nuclear, mitochondrial, and chloroplast genomes.arrow_forwardTHE MOLECULAR GENETICS OF CYSTIC FIBROSIS and of The following is the base sequence of DNA that codes for amino acids 506-510 of the protein that regulates the chlorine channels in the cell membrane. This protein contains a total of 1476 amino acids so this is a small part of the entire gene. DNA Template Strand: 3'TAGTAGAAACCACAA5' 1. What is the minimum number of DNA nucleotides in this whole gene? 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 3. What mRNA will be formed from the template strand of DNA? 4. What amino acids will this mRNA code for? 5. If the 6th, 7th and 8th bases in the template strand of the DNA are removed, rewrite the new template strand below. 6. When the template strand of the DNA is changed, this is referred to as a mutation. What kind of mutation is this? 65arrow_forwardFig 2 shows gastric organoids cultured from a TffCre; XYZI/fl mouse. Ki67 has been used to identify proliferating cells (green). Tamoxifen indicates exposure to tamoxifen to induce recombination of the TffCre. No Tamoxifen Ki-67 3. What is the consequence of deleting XYZ in gastric organoids, and does this suggest it is an oncogene or a tumour suppressor? Explain your reasoning. DX Equation Editor 4. IWP inhibits the secretion of Wnt ligands from the organoids. Therefore, do Wnt ligands promote or inhibit proliferation of XYZ deficient organoids? Explain your reasoning. X Equation Editor Tamoxifen Q Tamoxifen + IWP A- A I B IU S x, x² Styles Font Size Font Words: 0/250 张 A- A-T BIUS X, x² Styles ▾ Sizearrow_forward
- The human RefSeq of the entire first exon of a geneinvolved in Brugada syndrome (a cardiac disordercharacterized by an abnormal electrocardiogram andan increased risk of sudden heart failure) is:5′ CAACGCTTAGGATGTGCGGAGCCT 3′The genomic DNA of four people (1–4), three ofwhom have the disorder, was subjected to singlemolecule sequencing. The following sequences represent all those obtained from each person. Nucleotidesdifferent from the RefSeq are underlined. Individual 1:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGTGCGGAGACT 3′Individual 2:5′ CAACGCTTAGGATGTGAGGAGCCT 3′Individual 3:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGGCGGAGCCT 3′Individual 4:5′ CAACGCTTAGGATGTGCGGAGCCT 3′and5′ CAACGCTTAGGATGTGTGGAGCCT 3′a. The first exon of the RefSeq copy of this gene includes the start codon. Write as much of the aminoacid sequence of the encoded protein as possible,indicating the N-to-C polarity.b. Are any of these individuals homozygotes? If so,which person and what allele?c. Is…arrow_forwardRecall that Leber congenital amaurosis (LCA), a formof congenital blindness in humans, can be caused byhomozygosity for recessive mutations in the RPE65gene. Recently, a rare dominant mutation in RPE65has been implicated as one cause of an eye diseasecalled retinitis pigmentosa, which is characterized byretinal degeneration that can progress to blindness.The dominant RPE65 mutation is a missense mutation causing amino acid 447 in the polypeptide tochange from Asp to Glu. Little is known about thenature of the mutant protein.a. Do you think that the dominant allele is morelikely a loss-of-function or a gain-of-functionmutation? Explain.b. As described in this chapter, gene therapy for LCAhas been at least partially successful. Do you thinkthat the same kind of gene therapy can be used forpatients with retinitis pigmentosa caused by thedominant mutant allele of RPE65? Explain.arrow_forwardPeople who carry a theoretical genetic disorder (called B-disease) can be identified from a 2kb DNA sequence. People who carry this genetic disorder have a single nucleotide polymorphism that results in a change of GTATCC to GGATCC, a site that only occurs once at nucleotide number 750 in this DNA sequence. Answer the following questions based on the information provided. (a) How can you develop a simple molecular test to identify the genetic disorder?r B-dif w. (41 (b) If you have carried out the molecular test (based on the information above) on a 100 individual and found that 24 were healthy (BB) and 26 were carriers (bb); 1) What is the ratio of heterozygous? 2) Show how can you identify the three types from the agarose gel (H focaiarrow_forward
- In human DNA, 70% of cytosine residues that are followed by guanine (so-called CpG dinucleotides, where p indicates the phosphate in the phosphodiester bond between these two nucleotides) are methylated to form 5-methylcytosine. As shown in the following figure, if 5-methylcytosine should undergo spontaneous deamination, it becomes thymine. Methylated CpG dinucleotides are hotspots for point mutations in human DNA. Can you propose a hypothesis that explains why?arrow_forwardIn the human genome for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotide in the amino acid coding region is represented by the sequence 3'-TACCACGTGGACTGAGGACTCCTCTTCAGA-5'arrow_forwardThe D1S80 locus is located on human chromosome 1 and is characterized by a repeating 16 base pair (bp) sequence. Alleles for this locus vary depending on the number of repeats present, thus affecting the size of the locus. The D1S80 locus also contains two conserved sequences, a 32bp sequence at one end and a 113bp sequence at the other end. If the DNA of an individual is targeted for D1S80 amplification, and one of the resulting amplicons is approximately 785bp in size, how many repeats would be present in this D1S80 allele? The amplicon of interest is indicated by a red arrow in the diagram below.arrow_forward
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