Genetics Dideoxy Sequencing Notes

Describe and list examples of Dideoxy sequencing Description: -creates set of DNA strands of different lengths -color coded -more for small laboratory experiments -sequences one fragment at a time -Computers are used to read the frequences (also in next-generation sequencing) -The computer detects the peaks to determine the sequencing -Based on DNA polymerase (also in next-generation sequencing) Examples: -reaction produces set of DNA strands of different lengths -Used to determine sequence of single DNA fragment Describe and list examples of next-generation sequencing Description: -all produces similar fragment lengths -able to determine the sequence of the whole, or large part of the genome -produce large amounts of data -sequences many fragments at a time -Computers are used to read the frequences (also in Dideoxy sequencing) -The massive amounts of data are processed through a computer -Based on DNA polymerase (also in Dideoxy sequencing) Examples: -suitable for personal genomics applications - Many simultaneous reactions Suppose a scientist wants to clone the gene for human insulin. He uses a plasmid that contains the lacZ gene, an origin of replication, and an ampicillin resistant gene. He inserts the human insulin gene into a restriction site located within the lacZ gene. Then he transforms this plasmid into bacteria without a functional lacZ gene and spreads the appropriate dilution onto bacterial plates. When the product of the lacZ gene, beta-galactosidase, interacts with the substrate X-Gal on LB agar, colonies turn blue. Which plate indicates that the human insulin gene was successfully inserted in the plasmid? Answer: -an LB agar plate with ampicillin and X-Gal that produces colorless colonies Explanation: -origin of replication is in all plasmids -ampicillin resistance is for selection to have bacterial cells that receive the plasmid -lacZ gene is used for a color reaction to determine which plasmids received an insert -when lacZ gene is intact and can produce product (is white) -Because it is blue, that means the plasmid is not intact and cannot carry out the rest of its functions -look for interruption of the lacZ gene The table shows where different restriction enzymes cleave DNA. The abbreviation R represents the purines (adenine and guanine). The pyramids (cytosine, thymine, and uracil) are abbreviated as Y. The abbreviation W represents adenine or thymine. (table here) Which restriction endonucleases would cleave a DNA molecule at the given sequences? The complementary DNA substrate strand is omitted for clarity. Answer: -EcoRV Explanation: - Sequence in question to be cut: 5' GAAGATATCCTCGC 3' -EcoRV will cut (5' GATATC 3') -The cleavage column did not determine whether or not an enzyme would cut the sequence Researchers construct different types of genomic maps to gain information about genes and their locations in the genome. Why would a researcher choose to construct a physical map of genes rather than a genetic map? Answer: -To determine the locations of specific genes on a chromosome and the distance between them Explanation: -genetic maps indicate the relative position of genes as they are aligned along the chromosome, but does not give information about the space between the genes -physical maps are based on the sequence of the chromosome and give information on all the base pairs but may not give information on which base pairs are genes -don't indicate geographic locations of human populations -more than markers needed here -map does not give special information on their sequence such as height or weight green= intestine; gene D red= stomach; gene C Genes A and B are responsible for basic cellular functions and are equally expressed in stomach and intestinal tissues. Answer: Gene C=red Gene D=green Gene A and B = yellow As depicted in the table, certain plant genomes are significantly larger than mammal genomes, including the human genome. Why do these diploid plants have more DNA, but less organismal complexity compared with humans? Answer: -Repetitive DNA from transposons accounts for the majority of these plant genomes Explanation: -Most eukaryotes have about the same number of genes -some plants have 25 or 30 thousand genes and human had around 20 thousand genes -restate the question: (how can the genome be larger without many more genes?) -transposons can make up a large portion of genomes. They make up over 40% of the human genome. - transposons are not important in gene expression Some traits are considered to be quantitative even though phenotypes are classified only as present or absent. Genetic diseases often fit this description. What is the term used to describe this type of trait? Answer: -Threshold trait Explanation: -quantitative because there are many genes involved and they can be affected by the environment -Polygenic trait: multiple genes are underlying the phenotype -environmental trait: environment affects a trait, but this isn't actually a real term -meristic trait: only fixed values -epigenetic trait: controlled by the changes in genes other than changes in gene sequence -threshold trait: quantitative trait where the phenotype either shows are doesn't, but you need to accumulate enough factors (genetic, environmental) Some traits are considered to be quantitative even though phenotypes are classified only as present or absent. Genetic diseases often fit this description. What creates the underlying quantitative distribution for this type of trait? Anwer: -Genetic and environmental factors determine an individual's susceptibility to the trait. Susceptibility exhibits continuous variation, and individuals above a certain threshold express the trait Explanation: -Why doesn't everyone with a certain allele show the certain phenotype? -You need a certain collection of alleles that increases the susceptibility above a certain level -quantitative traits are in many loci, not a single locus Random sample values, Height values. Be able to calculate the mean and variance of the data. (not standard deviation) ... A breeder is breeding dogs to select for weight. The weight, in pounds, for the eight dogs in the population is shown in the chart. The mean weight of the population of dogs is 60 pounds. Classify each mating pair according to the effect of its offspring has on the mean weight of the population. A 40 B 60 C 60 D 80 E 40 F 60 G 80 H 60 Anwer: -Increase: C and D -Stays the same: B and H, C and F -Decrease: A and E, E and a 30 pound dog Explanation: -anything above 60 will increase, and anything below 60 will decrease the mean weight of the population Suppose that seed size in a particular plant species is a polygenic trait. A grower crosses two different pure-breeding varieties of plant, and measures seed size in the F1 progeny. She then backcrosses the F1 plants to one of the parental varieties and measures seed size in the backcross progeny. The grower finds that seed size in the backcross progeny has a higher variance than seed size in the F1 progeny. What is the most likely explanation for the higher seed size variability in the backcross progeny? Answer: -The backcross progeny are not uniformly heterozygous at the seed size loci Explanation: -Polygenic: multiple genes involved -Any variance that occurs in pure-breeding is due to environmental factors -If you have an F1 that is heterozygous for all those genes, many different combinations will segregate in the backcross -There are more genotypes with the backcross of the heterozygous -the backcross progeny are not uniformly heterozygous at all seed size loci (this is actually the F1 progeny) -cell divisions and mutations were not discussed in the heritability section Classify each statement about the heritability of a phenotype according to the narrow-sense heritability value it describes Answer: -0.02 (va is small): -Genes do not likely contribute to the phenotypic variability in the population -A relationship between the phenotypes of parents and their offspring is mostly determined by environmental factors -0.5 (va is about half): -half of the observed variation is due to the additive genetic effects of alleles -0.98 (va is almost all): - Parents and their offspring share heritable phenotypes -Phenotypic differences result from genetic variation Explanation: -(va/vp) Suppose a farmer is interested in developing a breeding program on his chicken farm. The farmer would like to artificially select for egg eight, egg shape, shell color, and shell thickness. In a large population of his chickens, the farmer has measured these four continuous egg traits of interest and calculated their variances, which are shown in the table. (Narrow-sense heritability) Which trait would best respond to artificial selection by the farmer? Answer: -Shell color (59/112= .53) Explanation: -The table gives the calculated variances, which gives us the heritability -under conditions of selection the response to selection R is R=H^2 x S -Response = heritability x selection differential -additive genetic and phenotype columns are very important -The answer is the trait with the highest H^2 -Additive genetic variance / phenotypic variance= answer -Weight (12/503=.02) -Shape (100/745=.13) -Shell thickness (20/736=.02) Suppose a farmer is interested in developing a breeding program on his chicken farm. The farmer would like to artificially select for egg eight, egg shape, shell color, and shell thickness. In a large population of his chickens, the farmer has measured these four continuous egg traits of interest and calculated their variances, which are shown in the table. (Narrow-sense heritability) Which trait would best respond to artificial selection by the farmer? A: Shell color (59/112= .53) Select all of the statements that identify how the farmer might utilize the information gained from his data. Answer: -Identify quantitative trait loci correlated to each trait examined in the chicken population after obtaining genetic sequence data for his chicken population. (Eventually. Identify IQTL if he has a mapping population, and genetic sequence data to map the phenotype of the different individuals) -Design an experiment to improve the shell thickness trait in the current chicken population by changing the farm environment (reducing the environmental variance) Explanation: -Cannot: -Recommend a selective breeding program for a population of chickens from another country based on his results (another country is another environment) Hardy-Weinberg calculations ... allele frequency from genotype frequency ... affects of migration (Mutation, migration, natural selection, and genetic drift) ... Change of evolution within a population ... Which of the techniques are examples of biotechnology? Answer: -Using DNA fragments to harvest insulin for the treatment of diabetes -breeding different dog breeds for hinting or racing -using yeast to ferment fruits or grains to make beer, wine, whiskey, or vodka -developing transgenic crops to use as biofuels and reduce carbon emissions Explanation: -Biotechnology: any innovation in the use of biological resources like plants, animals, or microbes to make products of use to humans. -Biotechnology includes any use of biological resources to make a product of use for humans -Biotechnological techniques include both modern ones and ancient ones A geneticist isolates a new restriction enzyme from the bacterium aeromonas ranidae. No other restriction enzymes have been isolated from this bacterial species. Use the standard convention for abbreviating restriction enzymes to name this new restriction enzyme. Answer: -Ara1 Explanation: -The first three letters of a restriction enzyme name refer to the bacterial genus and species. A Roman numeral following the letters distinguishes multiple restriction enzymes isolated from the same species. -Example: "Eco" refers to E. Coli. A fourth letter may refer to the bacterial strain from which the enzyme was isolated. For example, the "R" in EcoRI indicates that this enzyme was isolated from E. coli strain RY13. -Roman numerals following the first three or four letters identify different enzymes isolated from the same species. The first enzyme to be isolated from a species would receive the Roman numeral 1. For both questions, assume that the four types of nucleotides are equally likely to be found in DNA. Calculate how often, on average, a type II restriction endonuclease is expected to cut a DNA molecule if the recognition sequence for the enzyme has 5 bp. How bout 8bp? Answer: -5: Once in every 1024 bp -> (4^5) -8: Once in every 65536 bp -> (4^8) Explanation: -Consider that each position in a DNA sequence is an independent variable when compared to each other position. Accordingly, the entire sequence will be multiplicative. -A restriction endonuclease that recognizes a specific sequence that is 5 bp long will cut once whenever the sequence appears. Consider that there are 4 possible nucleotides in a DNA sequence, and each base is independent of the other bases. -Expressed mathematically, a sequence of n bases will appear 1/(4^n) of the time. In other words, the sequence will appear once in every 4^n bases. Samples of a plasmid containing a segment of unknown DNA are digested using the restriction enzymes EcoRI, BamHI, and a combination of EcoRI and BamHI. The digests are then run on an agarose gel in order to separate the resulting fragments by size. Use the results of the gel electrophoresis shown in the image to determine the sizes of the fragments and label the restriction map. Create a map of the DNA segment by dragging the enzyme name to the location it cuts (the vertical lines above the segment) and labeling the fragments with their lengths (numbers, in kilobases). You will not use all the numbers. Answer: (map image) Match each vector type to the DNA fragment size that vector may clone. Answer: -4 kb = plasmid -20 kb = bacteriophage -35 kb = cosmid -100 kb = bacterial artificial chromosome (BAC) Explanation: -Different types of vectors accommodate different sizes of DNA inserts. These vectors derive from natural sources. Scientists engineer plasmid vectors from natural bacterial plasmids, or small pieces of DNA that replicate independently of larger bacterial chromosomes. -Bacteriophage can only clone DNA fragment sizes up to 15 kb Describe the construction of a recombinant plasmid containing the gene for the green fluorescent protein (GFP) and the insertion of the plasmid into a bacterial cell by placing the steps in order. Answer: -Use PCR to amplify the gene for GFP -Perform a restriction digestion of the GFP gene and the plasmid -Ligate together the GFP gene and the plasmid to generate a recombinant plasmid -Transform bacteria with the recombinant plasmid using electroporation -Plate the bacterial cells, and screen for positive transformants Explanation: -The first step to construct the recombinant plasmid involves making many copies of the green fluorescent protein (GFP) gene. A recombinant plasmid is a circular piece of DNA that includes the plasmid backbone and the gene of interest. A geneticist uses a cloning plasmid that contains the lacZ gene and a gene that confers resistance to penicillin. She inserts a piece of foreign DNA into a restriction site located within the lacZ gene and uses the plasmid to transform bacteria. She then grows the bacteria on selective media containing penicillin and X-gal. Explain how the geneticist can identify bacteria that contain a plasmid with the foreign DNA. Answer: -Bacteria with the desired plasmid produce white colonies Explanation: -Selective media allows a scientist to screen for bacterial cells containing a desired plasmid. Penicillin ensures that only cells containing the plasmid that confers antibiotic resistance will form colonies. X-gal discriminates between cells with or without functioning lacZ genes. -Blue = unsuccessful plasmid/gene transfer -White = successful plasmid/gene transfer Plasmids are small circular DNA molecules found in bacteria that replicate separately from chromosomes. Why are plasmids essential for recombinant DNA technology? Answer: -DNA from a gene of interest can be inserted into a plasmid, then the modified plasmid can be inserted into a bacterial cell to replicate a gene of interest many times Explanation: -Plasmids serve as vectors for the generation of recombinant DNA. A vector is a small piece of DNA in which foreign DNA can be inserted and stably maintained when the combination is inserted into a cell or organism. A researcher treats a DNA sample and a plasmid with the ampR and lacZ (for the enzyme β-galactosidase) genes with the restriction enzyme EcoRI with the goal of making recombinant DNA molecules. The researcher places the lacZ gene in the cloning region so that it is disrupted by DNA inserts. Place the remaining steps for this process in the correct order. Not all steps will be used. Answer: 1) All the fragments have identical sticky ends: 5' -AATT- 3' 2) Use DNA ligase to join the sticky ends of the DNA sample and the plasmid 3) Grow cells containing the recombinant plasmid in a medium with ampicillin and X-gal (a galactose derivative) 4) White colonies that survive on the culture medium contain the recombinant plasmid 5) Remove colonies containing the recombinant plasmid from the culture medium and continue growing Explanation: -The restriction enzyme EcoRI recognizes and cleaves the sequence G ∣∣ AATTCG | AATTC . Consider the sticky ends that result from this cleavage. -The ampR gene confers resistance to the antibiotic ampicillin. -The lacZ gene codes for β-galactosidase, which hydrolyzes some galactose derivatives. Cells that contain an intact lacZ gene hydrolyze X-gal, which yields a blue product. Inserted DNA can interrupt this gene. Consider the color of colonies that do not have an intact lacZ gene. Suppose a scientist is studying the platypus, one of the few venomous mammals. The male platypus produces venom in the crural glands located beneath the calcaneus spur on each of its hind limbs. The female platypus can develop calcaneous spurs, but the crural glands never develop. Which of the following libraries would be most useful to help the scientist identify the gene or genes responsible for venom production? Answer: -cDNA library made from the mRNA isolated from the crural gland of a male platypus Explanation: -Remember that cDNA libraries are generated from isolated mRNAs, whereas genomic DNA libraries contain all the sequence from the organism's genome. -The genomic library will be the same regardless of the tissue from which it is isolated -Venom is comprised of a number of different proteins, so the scientist needs to look for genes that are expressed -scientist needs a library that can identify genes expressed in a specific tissue Complementary DNA (cDNA) is a double-stranded molecule. In the laboratory, how is cDNA generated from a eukaryotic messenger RNA (mRNA)? Answer: -Reverse transcriptase generates a single-stranded cDNA and then DNA polymerase synthesizes the complementary strand Explanation: -Reverse transcriptase can synthesize a single-stranded DNA molecule from a single-stranded eukaryotic mRNA. What enzyme that is used within the cell to synthesize new DNA from a DNA template can be used to generate the complementary strand? -cDNA: double-stranded DNA molecule that is a copy of a messenger RNA molecule -cDNA can be generated from mRNA using an enzyme called reverse transcriptase Polymerase chain reaction (PCR) is a technique used to amplify (copy) DNA. Suppose a single, linear molecule of double-stranded DNA (dsDNA) is amplified by PCR. After one PCR cycle, how many molecules of dsDNA will there be? After 3? Answer: - 2 molecules of dsDNA after one cycle -8 molecules of dsDNA after three cycles Explanation: -Template DNA is incubated with primers, nucleotides, and a thermostable polymerase in a buffer with Mg2+Mg2+ ions (required for polymerase activity). -PCR reaction: -There is an initial denaturation step at 95 °C and then the following steps are cycled 25-30 times. 1) 95 °C - Denaturation Time - to generate ssDNA 2) 55 °C - Annealing Time - primers bind to complementary regions of the template DNA 3) 68 °C - Extension Time - the polymerase extends the primers to form dsDNA.