Consider the problem of searching for genes in DNA sequences using Hor-spool’s algorithm. ADNA sequence is represented by a text on the alphabet {A, C, G, T}, and the gene or gene segment is the pattern.  a. Construct the shift table for the following gene segment of your chromo-some 10: TCCTATTCTT  b. Apply Horspool’s algorithm to locate the above pattern in the following DNA sequence: TTATAGATCTCGTATTCTTTTATAGATCTCCTATTCTT

We are given a random DNA sequence (each of the 4 bases has equal probability of occurring at each position in the sequence). A. What is the probability that three adjacent bases (a triplet) encode a stop codon? B. What is the probability that a given triplet is not a stop codon? C. What is the probability that neither of two adjacent triplets is a stop codon? D. What is the probability that none of 100 adjacent triplets are stop codons? E. Consider a single base insertion mutation between the 3rd and 4th codons in a natural gene that encodes a protein 100 amino acids long. Is it more likely that the protein produced by this mutant allele will be shorter or longer than 100 amino acids long?

Transcriptome analysis involves two separate methodologies: gene expression and RNA seq analyses. The 10 items below are a scrambled listing of the steps used in the two procedures. Identify the steps involved in RNA seq from the list below. Use the numbers in the list to refer to each step. Once the steps for RNA seq have been identified, write the steps in the order in which they are performed during the experiment.   (1) DNA sequencing (2) Allow for hybridization and wash excess cRNA. (3) Mix labeled cRNA with array chip. (4) PCR amplification (5) Measure fluorescence intensity to determine abundance of transcripts. (6) Add labeled cRNA at each microarray location. (7) Map cDNA sequences to the genome of the organism to determine identity and abundance of transcripts. (8) mRNA isolation from cells (9) Prepare fluorescently labeled cRNA probes (10) cDNA synthesis

Answer part C of the following: A) E. coli DNA and binturong DNA are both 50% G-C. If you randomly shear E. coli DNA into 1000 bp fragments and put it through density gradient equilibrium centrifugation, you will find that all the DNA bands at the same place in the gradient, and if you graph the distribution of DNA fragments in the gradient you will get a single peak (see below). If you perform the same experiment with binturong DNA, you will find that a small fraction of the DNA fragments band separately in the gradient (at a different density) and give rise to a small "satellite" peak on a graph of the distribution of DNA fragments in the gradient (see below). Why do these two DNA samples give different results, when they're both 50% G-C? See graph B.    B) If you denatured the random 1000 bp fragments of binturong DNA that you produced in question 1a by heating them to 95ºC, and then cooled them down to 60ºC and allowed them to reanneal, you would find that approximately 15% of the…