BL414 Genetics Spring 2006 page Test 2
... 1) (2.5pts) T or F: ___T_______ Bacterial genomes do not have many repetitive sequences, most of their genome is unique. 2) (2.5pts) T or F: ____F______ Genetic linkage in corn can be analyzed using asci tetrad analysis. 3) (2.5pts) T or F: _____F_____ The Holliday model is the currently accepted mo ...
... 1) (2.5pts) T or F: ___T_______ Bacterial genomes do not have many repetitive sequences, most of their genome is unique. 2) (2.5pts) T or F: ____F______ Genetic linkage in corn can be analyzed using asci tetrad analysis. 3) (2.5pts) T or F: _____F_____ The Holliday model is the currently accepted mo ...
5.DNA - Colorado State University
... single cell to discover the identity of individuals who have perpetrated crimes. Some sequences of DNA are exactly like the sequences found in bacteria. These are codes for very important organelles that make the proteins. Other sequences, that don’t code for anything (called “junk DNA”), can mutate ...
... single cell to discover the identity of individuals who have perpetrated crimes. Some sequences of DNA are exactly like the sequences found in bacteria. These are codes for very important organelles that make the proteins. Other sequences, that don’t code for anything (called “junk DNA”), can mutate ...
Nucleotides - Mrs Miller's Blog | Science Revision
... • The chains are always the same distance apart because bases pair up in a specific way. • When a purine appears on one side, a pyrimidine appears on the other • Adenine always pairs with thymine, and guanine always pairs with cytosine • As the strands come together, hydrogen bonds form between the ...
... • The chains are always the same distance apart because bases pair up in a specific way. • When a purine appears on one side, a pyrimidine appears on the other • Adenine always pairs with thymine, and guanine always pairs with cytosine • As the strands come together, hydrogen bonds form between the ...
Bio 93 2013 Final: 1. Which option best describes transformation in
... C) Segment 1 would denature at a lower temperature than segment 2 D) Segment 2 would denature at a lower temperature than segment 1 10. The leading and the lagging strands differ in that A) the leading strand is synthesized by ...
... C) Segment 1 would denature at a lower temperature than segment 2 D) Segment 2 would denature at a lower temperature than segment 1 10. The leading and the lagging strands differ in that A) the leading strand is synthesized by ...
Deoxyribonucleic Acid (DNA)
... DNA and Chromosomes • __________: DNA located in _______ in form of ___________. – Number _______ widely between _______. – Humans = 46 chromosomes (23 pairs). ...
... DNA and Chromosomes • __________: DNA located in _______ in form of ___________. – Number _______ widely between _______. – Humans = 46 chromosomes (23 pairs). ...
DNA Review
... Chain Reaction (PCR) that is the base of forensic DNA analysis D. In vivo DNA synthesis: 1. Replication in eukaryotes starts at many points of origin and spreads with many replication bubbles—places where the DNA strands are separating. 2. It needs many more factors, unlike the replication in vitro. ...
... Chain Reaction (PCR) that is the base of forensic DNA analysis D. In vivo DNA synthesis: 1. Replication in eukaryotes starts at many points of origin and spreads with many replication bubbles—places where the DNA strands are separating. 2. It needs many more factors, unlike the replication in vitro. ...
Escherichia coli
... Figure 14.5. Replication slippage. The diagram shows replication of a five-unit CA repeat microsatellite. Slippage has occurred during replication of the parent molecule, inserting an additional repeat unit into the newly synthesized polynucleotide of one of the daughter molecules. When this daught ...
... Figure 14.5. Replication slippage. The diagram shows replication of a five-unit CA repeat microsatellite. Slippage has occurred during replication of the parent molecule, inserting an additional repeat unit into the newly synthesized polynucleotide of one of the daughter molecules. When this daught ...
pptx - WVU School of Medicine
... DNA sequences “upstream” of transcription initiation site. • different σ factors recognize different promoters (σ70 = most genes; σ32 = heat shock proteins; σ28 = flagella & chemotaxis genes). • 2 DNA sequences (-35 & -10) found in most prokaryotic promoters – “upstream” of transcription start site ...
... DNA sequences “upstream” of transcription initiation site. • different σ factors recognize different promoters (σ70 = most genes; σ32 = heat shock proteins; σ28 = flagella & chemotaxis genes). • 2 DNA sequences (-35 & -10) found in most prokaryotic promoters – “upstream” of transcription start site ...
GEL ELECTROPHORESIS VIRTUAL LAB
... For each section read the question first and then read through the information on the website. As you go through the virtual lab, be sure to read all directions, follow all prompts given to you, and answer all of the following questions. DNA STRAND SIZE ...
... For each section read the question first and then read through the information on the website. As you go through the virtual lab, be sure to read all directions, follow all prompts given to you, and answer all of the following questions. DNA STRAND SIZE ...
Biology 303 EXAM II 3/14/00 NAME
... 1. leads to the condition of “semisterility” even in the absence of any crossing-over. 2. leads to semisterility only if a crossover occurs between the translocated chromosomes during meiosis. 3. greatly increases the chances of nondisjunction. 4. has no consequence since the genetic material is sti ...
... 1. leads to the condition of “semisterility” even in the absence of any crossing-over. 2. leads to semisterility only if a crossover occurs between the translocated chromosomes during meiosis. 3. greatly increases the chances of nondisjunction. 4. has no consequence since the genetic material is sti ...
Biology 303 EXAM II 3/14/00 NAME
... 1. leads to the condition of “semisterility” even in the absence of any crossing-over. 2. leads to semisterility only if a crossover occurs between the translocated chromosomes during meiosis. 3. greatly increases the chances of nondisjunction. 4. has no consequence since the genetic material is sti ...
... 1. leads to the condition of “semisterility” even in the absence of any crossing-over. 2. leads to semisterility only if a crossover occurs between the translocated chromosomes during meiosis. 3. greatly increases the chances of nondisjunction. 4. has no consequence since the genetic material is sti ...
REPLICATION A DNA molecule separates into two template strands
... of DNA that are required for various experiments and applications in molecular biology, forensic analysis, evolutionary biology, and medical diagnostics. The PCR technique is based on the natural processes a cell uses to replicate a new DNA strand. Only a few biological ingredients are needed for PC ...
... of DNA that are required for various experiments and applications in molecular biology, forensic analysis, evolutionary biology, and medical diagnostics. The PCR technique is based on the natural processes a cell uses to replicate a new DNA strand. Only a few biological ingredients are needed for PC ...
Learning Guide: Molecules of Life Bill Activity #19 1st Read About
... o List the monomer of carbohydrates and its basic formula. List the two functional groups that all sugars have. o Sketch the abbreviated ring structure of glucose. Number the carbons on the structure. Glucose is a monosaccharide. Describe the important function of these molecules. o Explain/sketch h ...
... o List the monomer of carbohydrates and its basic formula. List the two functional groups that all sugars have. o Sketch the abbreviated ring structure of glucose. Number the carbons on the structure. Glucose is a monosaccharide. Describe the important function of these molecules. o Explain/sketch h ...
Bio1001Ch12W
... form hydrogen bonds, connecting the two strands. • Adenine could form two ________ bonds only with thymine • Guanine could form ______ hydrogen bonds only with cytosine. • This finding _________ Chargaff’s rules. Fig. 16.6 ...
... form hydrogen bonds, connecting the two strands. • Adenine could form two ________ bonds only with thymine • Guanine could form ______ hydrogen bonds only with cytosine. • This finding _________ Chargaff’s rules. Fig. 16.6 ...
Scientist Guide DNA Bracelet Workshop
... uses a blueprint to construct a house, cells use DNA to construct an organism. DNA is therefore often considered the “blueprint for life.” The DNA instructions are divided into segments called genes. All organisms have genes that determine various biological traits, some of which are immediately vis ...
... uses a blueprint to construct a house, cells use DNA to construct an organism. DNA is therefore often considered the “blueprint for life.” The DNA instructions are divided into segments called genes. All organisms have genes that determine various biological traits, some of which are immediately vis ...
DNA Replication and DNA Repair Study Guide Focus on the
... b. Discontinuous. i. WƌŽĐĞĞĚƐŝŶϱ͛ƚŽϯ͛ĚŝƌĞĐƚŝŽŶ;ŽŶϯ͛ƚŽϱ͛ƐƚƌĂŶĚͿ ii. Very fragmented 5. Role of helicases, topoisomerases,and single stranded binding proteins a. Helicases i. Unwind DNA ii. Require ATP 1. Hydrolyzed in order to function iii. Supercoiling-increased or decreased torsional stra ...
... b. Discontinuous. i. WƌŽĐĞĞĚƐŝŶϱ͛ƚŽϯ͛ĚŝƌĞĐƚŝŽŶ;ŽŶϯ͛ƚŽϱ͛ƐƚƌĂŶĚͿ ii. Very fragmented 5. Role of helicases, topoisomerases,and single stranded binding proteins a. Helicases i. Unwind DNA ii. Require ATP 1. Hydrolyzed in order to function iii. Supercoiling-increased or decreased torsional stra ...
Towards DNA sequencing by force
... •We have inferred DNA thermodynamics using optical tweezers and performing single molecule experiments. •The NN model is useful to extract information about the intermediate states from the experimental noise measurements. •Sequencing DNA by force is not possible yet •Cooperative avalanches (intrins ...
... •We have inferred DNA thermodynamics using optical tweezers and performing single molecule experiments. •The NN model is useful to extract information about the intermediate states from the experimental noise measurements. •Sequencing DNA by force is not possible yet •Cooperative avalanches (intrins ...
DNA
... diameter indicated by the X-ray data. – A purine-purine pair would be too wide and a pyrimidinepyrimidine pairing would be too short. – Only a pyrimidinepurine pairing would produce the 2-nm ...
... diameter indicated by the X-ray data. – A purine-purine pair would be too wide and a pyrimidinepyrimidine pairing would be too short. – Only a pyrimidinepurine pairing would produce the 2-nm ...
The DNA repair helicase UvrD is essential for replication
... Result is a long single-stranded gap on the leading strand side of the fork and the 5'-PO4 ended lagging strand "priming" the other side of the fork The 5'-PO4 ended "priming" strand is NOT a substrate for PriA so… for what is this a substrate? ...
... Result is a long single-stranded gap on the leading strand side of the fork and the 5'-PO4 ended lagging strand "priming" the other side of the fork The 5'-PO4 ended "priming" strand is NOT a substrate for PriA so… for what is this a substrate? ...
replication
... you have ALL your DNA in ALL your cells. If your cell needs to divide, you must have the DNA in there! ...
... you have ALL your DNA in ALL your cells. If your cell needs to divide, you must have the DNA in there! ...
Exam II Notes DNA
... When you make gametes, you can only have 23 chromosomes, not 46. You want one of each homologous chromosome, not two of each chromosome. (Why? Because if the egg has 23 chromosomes and the sperm has 23 chromosomes, then the fertilized egg will have 46 chromosomes, two of each number!) As mentioned a ...
... When you make gametes, you can only have 23 chromosomes, not 46. You want one of each homologous chromosome, not two of each chromosome. (Why? Because if the egg has 23 chromosomes and the sperm has 23 chromosomes, then the fertilized egg will have 46 chromosomes, two of each number!) As mentioned a ...
Topic 3 The chemistry of life
... 48. Helicase is the enzyme that breaks the hydrogen bonds to allow the unwinding. 49. The exposed bases of each strand are then paired with an available nucleotide by complementary base pairing. The result is two strands where only one was first present. 50. DNA polymerase is an enzyme that allows t ...
... 48. Helicase is the enzyme that breaks the hydrogen bonds to allow the unwinding. 49. The exposed bases of each strand are then paired with an available nucleotide by complementary base pairing. The result is two strands where only one was first present. 50. DNA polymerase is an enzyme that allows t ...
2.7 Review - Peoria Public Schools
... 48. Helicase is the enzyme that breaks the hydrogen bonds to allow the unwinding. 49. The exposed bases of each strand are then paired with an available nucleotide by complementary base pairing. The result is two strands where only one was first present. 50. DNA polymerase is an enzyme that allows t ...
... 48. Helicase is the enzyme that breaks the hydrogen bonds to allow the unwinding. 49. The exposed bases of each strand are then paired with an available nucleotide by complementary base pairing. The result is two strands where only one was first present. 50. DNA polymerase is an enzyme that allows t ...
DNA / RNA / PROTEIN SYNTHESIS / AP Biology
... mRNA Strand (from above) a. b. c. DNA Strand (from above) a. b. c. 6) Using your Amino Acids, mRNA from #5 above: a. What change (Amino Acids produced) would take place if an “A” was inserted before the first codon on “a” ? Write down the new Amino Acids from this mutation in the table below. What t ...
... mRNA Strand (from above) a. b. c. DNA Strand (from above) a. b. c. 6) Using your Amino Acids, mRNA from #5 above: a. What change (Amino Acids produced) would take place if an “A” was inserted before the first codon on “a” ? Write down the new Amino Acids from this mutation in the table below. What t ...
Molecular Biology
... • The alphabet of RNA is A, U, G and C • Within a molecule of mRNA, groups of 3 sequential nucleotides form meaningful “words” called codons – complementary to triplets in the template strand of the gene that was transcribed by RNA polymerase • each codon is a code for an amino acid of the protein c ...
... • The alphabet of RNA is A, U, G and C • Within a molecule of mRNA, groups of 3 sequential nucleotides form meaningful “words” called codons – complementary to triplets in the template strand of the gene that was transcribed by RNA polymerase • each codon is a code for an amino acid of the protein c ...
Holliday junction
A Holliday junction is a branched nucleic acid structure that contains four double-stranded arms joined together. These arms may adopt one of several conformations depending on buffer salt concentrations and the sequence of nucleobases closest to the junction. The structure is named after the molecular biologist Robin Holliday, who proposed its existence in 1964.In biology, Holliday junctions are a key intermediate in many types of genetic recombination, as well as in double-strand break repair. These junctions usually have a symmetrical sequence and are thus mobile, meaning that the four individual arms may slide though the junction in a specific pattern that largely preserves base pairing. Additionally, four-arm junctions similar to Holliday junctions appear in some functional RNA molecules.Immobile Holliday junctions, with asymmetrical sequences that lock the strands in a specific position, were artificially created by scientists to study their structure as a model for natural Holliday junctions. These junctions also later found use as basic structural building blocks in DNA nanotechnology, where multiple Holliday junctions can be combined into specific designed geometries that provide molecules with a high degree of structural rigidity.