AP Biology Microbiology unit Viruses, Bacteria
... 14. What is the phage reproductive cycle that replicates the phage genome without destroying the host? 15. What are phages called that care capable of using both modes of reproduction? 16. What is a prophage? ...
... 14. What is the phage reproductive cycle that replicates the phage genome without destroying the host? 15. What are phages called that care capable of using both modes of reproduction? 16. What is a prophage? ...
Name - drpence
... 23. As cells become more complex, they lose more coverings. What does this allow them to do? ...
... 23. As cells become more complex, they lose more coverings. What does this allow them to do? ...
U - Lakewood City Schools
... END PRODUCT –THE PROTEIN! The end products of protein synthesis is a primary structure of a protein A sequence of amino acid bonded together by peptide bonds ...
... END PRODUCT –THE PROTEIN! The end products of protein synthesis is a primary structure of a protein A sequence of amino acid bonded together by peptide bonds ...
What does PCR stand for?
... Chromosome 16 either has this Alu element or it doesn’t Since chromosomes come in pairs, a person is either ++, +-, or -- . DNA sequences such as Alu are used in diagnosis of genetic disease, forensic identification, and paternity ...
... Chromosome 16 either has this Alu element or it doesn’t Since chromosomes come in pairs, a person is either ++, +-, or -- . DNA sequences such as Alu are used in diagnosis of genetic disease, forensic identification, and paternity ...
April 8
... Oleosomes: oil-storing organelles with only outer leaflet • Put oils between the leaflets as they are made • Add oleosin proteins to outside: curve the membrane • Oils often have unusual fatty acids ...
... Oleosomes: oil-storing organelles with only outer leaflet • Put oils between the leaflets as they are made • Add oleosin proteins to outside: curve the membrane • Oils often have unusual fatty acids ...
Fe2+ is absorbed from the lumen of the gut (in the small intestine) by
... D. (2 pts each answer). Suppose you can make a recombinant yeast gene that contains the first 50 codons for a secreted protein followed by the entire coding region for histone 4. D-1. The resulting recombinant protein should have both an NLS and a SP. D-2. The recombinant protein will probably end u ...
... D. (2 pts each answer). Suppose you can make a recombinant yeast gene that contains the first 50 codons for a secreted protein followed by the entire coding region for histone 4. D-1. The resulting recombinant protein should have both an NLS and a SP. D-2. The recombinant protein will probably end u ...
Bacterial Transformation and Green Fluorescent
... splicing. Recall that genes are pieces of DNA that provide the instructions for making proteins. Proteins are essential molecules for all organisms and cause the expression of a trait, or phenotype, which is associated with that gene. In this specific procedure, we will perform genetic transformatio ...
... splicing. Recall that genes are pieces of DNA that provide the instructions for making proteins. Proteins are essential molecules for all organisms and cause the expression of a trait, or phenotype, which is associated with that gene. In this specific procedure, we will perform genetic transformatio ...
Chloroplast DNA and Molecular Phylogeny
... extremely low over the chloroplast genome as a whole and in most sequenced genes3. In spite of the low overall rate of cpDNA sequence evolution, different chloroplast genes do evolve at different rates: thus providing a range of molecular yardsticks for measuring evolutionary distances at various ta ...
... extremely low over the chloroplast genome as a whole and in most sequenced genes3. In spite of the low overall rate of cpDNA sequence evolution, different chloroplast genes do evolve at different rates: thus providing a range of molecular yardsticks for measuring evolutionary distances at various ta ...
Geometrical ordering of DNA in bacteria
... Figure 1. Average subcellular position of genes as function of their position on the chromosome in C. crescentus and E. coli as obtained from numerical simulations of compacted DNA. (A and B) show the position of genes along the cell axis as function of their position on the chromosomal map for an a ...
... Figure 1. Average subcellular position of genes as function of their position on the chromosome in C. crescentus and E. coli as obtained from numerical simulations of compacted DNA. (A and B) show the position of genes along the cell axis as function of their position on the chromosomal map for an a ...
Document
... 9. Gram-___________ bacteria have lipopolysaccharides on their cell wall, meaning they are ___________ resistant to antibiotics, which prevent peptidoglycan cross-linking. a. Negative…more b. Negative…less c. Positive…less d. Positive…more 10. Bacterial cells, but not eukaryotic cells, possess _____ ...
... 9. Gram-___________ bacteria have lipopolysaccharides on their cell wall, meaning they are ___________ resistant to antibiotics, which prevent peptidoglycan cross-linking. a. Negative…more b. Negative…less c. Positive…less d. Positive…more 10. Bacterial cells, but not eukaryotic cells, possess _____ ...
Bacteria/Virus review
... process is called and what happens. Binary Fission: Asexual, Conjugation: Sexual 15. List the different phylums of bacteria and give an example of each. 16. What are the 2 kingdoms in bacteria? Archeabacteria, Eubacteria 17. Why did they split Monera into 2 kingdoms? Different characteristics: missi ...
... process is called and what happens. Binary Fission: Asexual, Conjugation: Sexual 15. List the different phylums of bacteria and give an example of each. 16. What are the 2 kingdoms in bacteria? Archeabacteria, Eubacteria 17. Why did they split Monera into 2 kingdoms? Different characteristics: missi ...
Document
... In 1928, Heitz defined the heterochromatin as regions of chromosomes that do not undergo cyclical changes in condensation during cell cycle as the other chromosome regions (euchromatin) do. Heterochromatin is not only allocyclic but also very poor of active genes, leading to define it as geneti ...
... In 1928, Heitz defined the heterochromatin as regions of chromosomes that do not undergo cyclical changes in condensation during cell cycle as the other chromosome regions (euchromatin) do. Heterochromatin is not only allocyclic but also very poor of active genes, leading to define it as geneti ...
Lecture#23 - Cloning genes by complementation
... We need to screen the equivalent of 5000 molecules to ensure a 99% chance of finding gene A+. 5000 bacterial clones can be produced easily and screened quickly on one Petri dish plate. Works well with bacterial or yeast hosts and plasmid vectors, but what about higher organisms? -> problems with lar ...
... We need to screen the equivalent of 5000 molecules to ensure a 99% chance of finding gene A+. 5000 bacterial clones can be produced easily and screened quickly on one Petri dish plate. Works well with bacterial or yeast hosts and plasmid vectors, but what about higher organisms? -> problems with lar ...
Introduction to Nucleic Acids Definitions By definition
... relatively speaking, that utilize standard curve preparation to determine the concentration of analyte in blood (the BUN in this instance). A standard curve is a plot of the absorbance at a specific wavelength (in nm) of light by a colored compound on the “Y” axis versus the concentration of analyte ...
... relatively speaking, that utilize standard curve preparation to determine the concentration of analyte in blood (the BUN in this instance). A standard curve is a plot of the absorbance at a specific wavelength (in nm) of light by a colored compound on the “Y” axis versus the concentration of analyte ...
DNA SEQUENCING (using an ABI automated sequencer)
... Determination of a DNA sequence is accomplished using one of two basic methods, and their derivations. Both methods were first described in 1977. The first method (Maxam and Gilbert 1977) is based on specific chemical degradation of the DNA. The DNA is first end-labeled using 3 5 s or 33PI followed ...
... Determination of a DNA sequence is accomplished using one of two basic methods, and their derivations. Both methods were first described in 1977. The first method (Maxam and Gilbert 1977) is based on specific chemical degradation of the DNA. The DNA is first end-labeled using 3 5 s or 33PI followed ...
Basic Genetics & Background on Genetic Testing
... and a nitrogen base A , G , T or C • There are 4 different nitrogen bases in DNA and they can vary from one nucleotide to the next • The alternating bases provide the CODE ...
... and a nitrogen base A , G , T or C • There are 4 different nitrogen bases in DNA and they can vary from one nucleotide to the next • The alternating bases provide the CODE ...
Prodigiosin Production in E. Coli
... have been over diluted (we had our samples suspended in 1ml of solution, when Dr. Schwekendiek noted 100µl was the usual dilution) - To rectify this, we concentrated our DNA in a Speed Vacuum Concentrator overnight - After running our samples through the Speed Vacuum Concentrator, we ran them throug ...
... have been over diluted (we had our samples suspended in 1ml of solution, when Dr. Schwekendiek noted 100µl was the usual dilution) - To rectify this, we concentrated our DNA in a Speed Vacuum Concentrator overnight - After running our samples through the Speed Vacuum Concentrator, we ran them throug ...
Lab 1 Artificial Selection The purpose of a particular investigation
... 1. The frequency of two alleles in a gene pool is 0.1 9(A) and 0 .81(a). What is the percentage in the population of heterozygous individuals? What is the percentage of homozygous recessives? Assume that the population is in Hardy-Weinberg equilibrium. 2. An allele W, for white wool, is dominant ove ...
... 1. The frequency of two alleles in a gene pool is 0.1 9(A) and 0 .81(a). What is the percentage in the population of heterozygous individuals? What is the percentage of homozygous recessives? Assume that the population is in Hardy-Weinberg equilibrium. 2. An allele W, for white wool, is dominant ove ...
Lab Review - Warren County Schools
... 1. The frequency of two alleles in a gene pool is 0.1 9(A) and 0 .81(a). What is the percentage in the population of heterozygous individuals? What is the percentage of homozygous recessives? Assume that the population is in Hardy-Weinberg equilibrium. 2. An allele W, for white wool, is dominant ove ...
... 1. The frequency of two alleles in a gene pool is 0.1 9(A) and 0 .81(a). What is the percentage in the population of heterozygous individuals? What is the percentage of homozygous recessives? Assume that the population is in Hardy-Weinberg equilibrium. 2. An allele W, for white wool, is dominant ove ...
Restriction Enzymes and Electrophoresis - Milton
... In a previous activity you extracted DNA from your cheek cells. DNA extraction is the first step towards DNA analysis. In order for DNA to be analyzed for the presence of certain genes the extracted DNA must be prepared, or “chopped up”, into pieces with proteins called restriction enzymes. These pi ...
... In a previous activity you extracted DNA from your cheek cells. DNA extraction is the first step towards DNA analysis. In order for DNA to be analyzed for the presence of certain genes the extracted DNA must be prepared, or “chopped up”, into pieces with proteins called restriction enzymes. These pi ...
File
... VULNERABILITY OF CHILDREN TO MYCOSES 38) GRISEOFULVIN TREATS ATHLETE’S FOOT 39) MAJORITY OF THE POPULATION CARRIES CANDIDA BUT ONLY SOME SUFFER FROM CANDIDIASIS. CANDIDA IS THE MOST OPPORTUNISTIC FUNGUS. 40) A ZYGOSPORE IS A FERTILIZATION PRODUCT IN PHYCOMYCES. IT IS THE ZYGOSPORE THAT UNDERGOES RED ...
... VULNERABILITY OF CHILDREN TO MYCOSES 38) GRISEOFULVIN TREATS ATHLETE’S FOOT 39) MAJORITY OF THE POPULATION CARRIES CANDIDA BUT ONLY SOME SUFFER FROM CANDIDIASIS. CANDIDA IS THE MOST OPPORTUNISTIC FUNGUS. 40) A ZYGOSPORE IS A FERTILIZATION PRODUCT IN PHYCOMYCES. IT IS THE ZYGOSPORE THAT UNDERGOES RED ...
Science and Prokaryotes
... 11. Gram-___________ bacteria have many lipopolysaccharides on their cell wall, meaning they are ___________ resistant to antibiotics which prevent peptidoglycan cross-linking. a. Negative…more b. Negative…less c. Positive…less d. Positive…more 12. Bacterial cells, but not eukaryotic cells, possess ...
... 11. Gram-___________ bacteria have many lipopolysaccharides on their cell wall, meaning they are ___________ resistant to antibiotics which prevent peptidoglycan cross-linking. a. Negative…more b. Negative…less c. Positive…less d. Positive…more 12. Bacterial cells, but not eukaryotic cells, possess ...
11-17-11 DNA Lecture - Kings County Criminal Bar Association
... product rule; must use haplotypes (combination of alleles observed at all tested loci) ...
... product rule; must use haplotypes (combination of alleles observed at all tested loci) ...
Extrachromosomal DNA
Extrachromosomal DNA is any DNA that is found outside of the nucleus of a cell. It is also referred to as extranuclear DNA or cytoplasmic DNA. Most DNA in an individual genome is found in chromosomes but DNA found outside of the nucleus also serves important biological functions.In prokaryotes, nonviral extrachromosomal DNA is primarily found in plasmids whereas in eukaryotes extrachromosomal DNA is primarily found in organelles. Mitochondrial DNA is a main source of this extrachromosomal DNA in eukaryotes. Extrachromosomal DNA is often used in research of replication because it is easy to identify and isolate.Extrachromosomal DNA was found to be structurally different from nuclear DNA. Cytoplasmic DNA is less methylated than DNA found within the nucleus. It was also confirmed that the sequences of cytoplasmic DNA was different from nuclear DNA in the same organism, showing that cytoplasmic DNAs are not simply fragments of nuclear DNA.In addition to DNA found outside of the nucleus in cells, infection of viral genomes also provides an example of extrachromosomal DNA.