DNA sometimes has chemical groups attached, called methyl groups, that affect gene expression. Suppose that, during each hour, first a fraction m of unmethylated locations on the DNA become methylated, and then a fraction u of methylated locations become unmethylated. Find a recursion for the fraction f of the DNA molecule that is methylated.
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hello i need help with this
DNA sometimes has chemical
groups attached, called methyl groups, that affect gene
expression. Suppose that, during each hour, first a fraction
m of unmethylated locations on the DNA become methylated,
and then a fraction u of methylated locations become unmethylated. Find a recursion for the fraction f of the
DNA molecule that is methylated.
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- In the following gel showing stained bands of the Alu insertion sequence, what is the genotype of individual 2? 941 bp 641 bp->>> 1 2 3 4 5 6 Homozygous for the 641 bp sequence that does not contain in the Alu insertion Heterozygous, containing one 941 bp sequence and one 641 bp sequence O Homozygous for the 941 bp sequence containing the Alu insertionIn the human gene for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotides in the amino-acid-coding region is represented by the sequence: 3'-TACCACGTGGACTGAGGACTCCTCTTCAGA-5'. What is the sequence of the partner strand? 4B. If the DNA duplex for the beta chain of haemoglobin above were transcribed from left to right, deduce the base sequence of the RNA in this coding region. 4C. In NOT more than 200 words, explain how eukaryotic RNA synthesized by RNA polymerase II is modified before leaving the nucleus?e. four-base, not overlapping4. An example of a portion of the T4 rIIB gene in whichCrick and Brenner had recombined one + and one −mutation is shown here. (The RNA-like strand of theDNA is shown.)wild type 5′ AAA AGT CCA TCA CTT AAT GCC 3′mutant 5′ AAA GTC CAT CAC TTA ATG GCC 3′a. Where are the + and − mutations in the mutant DNA?b. The double mutant produces wild-type plaques.What alterations in amino acids occurred in thisdouble mutant?c. How can you explain the fact that amino acids aredifferent in the double mutant than in the wild-typesequence, yet the phage has a wild-type phenotype?
- Homologous Recombination, Heteroduplex DNA, and Mismatch Repair Homologous recombination in E. coli leads to the formation of regions of heteroduplex DNA. By definition, such regions contain mismatched bases. Why doesn’t the mismatch repair system of E. coli eliminate these mismatches?In a standard procedire, when writing and reading base sequences for nucleic acids (both DNA and RNAs) always to specify base sequence in 5' > 3' direction unless otherwise directed 1. From the base sequence 5' A-T-G-C-C-A 3' in a DNA template strand, determine the base sequence in hnRNA synthesized from the DNA template strand 2. From the base sequence 5' T-A-A- C-C-T 3' in a DNA template strand, determine the base sequence in hnRNA synthesized from the DNA template strandSupercoiled DNA is slightly unwound compared to relaxed DNA and this enables it to assume a more compact structure with enhanced physical stability. Describe the enzymes that control the number of supercoils present in the E. coli chromosome. How much would you have to reduce the linking number to increase the number of supercoils by five?
- 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…Are you a hidden heterozygote? A PCR analysis (part2) Agarose gel electrophoresis and interpretation la: Several factors (including agarose gel concentration, time and current) affect migration of DNA fragments through the agarose gel. Briefly explain how each of these factors affects DNA migration. Agarose gel concentration: Time: Voltage: 1b: Do DNA fragments move towards the positive or negative end of the gel box? Explain your answer. 1c: What is the purpose of the Tris-Acetate-EDTA (TAE) buffer that the agarose gel is prepared with and submerged in for running? What would happen if you used water to prepare and run the gel instead of TAE buffer? 1d: If the student is homozygous for the brown allele, how many bands will they see in the lanes for the blue and brown allele samples? (circle one) Brown sample: 0 Blue sample: 1 2 more than two. 1 2 more than two. le: If the student is homozygous for the blue allele, how many bands will they see in the lanes for the blue and brown allele…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?
- What is the length in AA’s of the LilP protein? Assume fMet is NOT CLEAVED. Enter just the number, nothing else! Write out the sequence of the polypeptide in AA: use the three letter notation, e.g. Met-Ser-Pro- A lilP mutant called lilPXS is isolated that produces a truncated polypeptide of only 6 AA in length. Describe a single basepair DNA change that would lead to this truncated version of the protein. Multiple options are possible (100 words max.)Given the fact that 1 fg of DNA = 9.78 * 105base pairs (on average), you can convert the amount of DNA per cell to the lengthof DNA in numbers of base pairs. (a) Calculate the number of basepairs of DNA in the haploid yeast genome. Express your answer inmillions of base pairs (Mb), a standard unit for expressing genomesize. Show your work. (b) How many base pairs per minute weresynthesized during the S phase of these yeast cells?Design a pair of primers to amplify the entire length of the following 45 base pair sequence.Make each primer 14 bases long. Write the sequences of the primers in 5' to 3' order.(Hint: It will help for you to write out BOTH strands of the DNA sequence listed below.5'-GATGCCCGTTGGATAAATTGGGCGTCTAGAATCGGTCACACTTAG-3'