🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
HomeClass Notes1,200,000US530,000

BSCI 223 Lecture Notes - Lecture 11: Allolactose, Start Codon, Catabolism

53 views3 pages
School
University of Maryland
Department
Biological Sciences Program
Course
BSCI 223
Professor
Dr.Mc Iver

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Base, and ribose then nucleoide, then triphosphate deoxyribonucleoide (dgtp) Bundle of dna that are assisted by proteins. Plasmid dna- encode for various factors for transferring of informaion. Dna helicase unzips double helix by breaking hydrogen bonds between complementary base pairs. Single stranded binding proteins stabilize the separated strands at the replicaion forks. The promoter is there at -10, which allows the rna polymerase to atached with the sigma factor- occur in 5" to 3". Rna polymerase can separate out the stand and then create the bubble and then synthesize an rna transcript. By breaking the diphosphate of, it provides every and uses the template dna strand for all nucleoides. Terminaion occurs because of road block- see above notes. Operon- single promoter upstream of muliple coding genes. The lac operon is negaively regulated by a transcripion factor. This transcripion factor is the laci gene product, the lac repressor (inhibitor) The lac repressor sterically inhibits rna polymerase from binding- blocking rna polymerase.

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers
Class+
$8 USD/m
Billed $96 USD annually
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
30 Verified Answers

Related Documents

BSCI 223 Lecture Notes - Lecture 9: Start Codon, Protein Production, Amine
maroonarmadillo314
BMS 501 Lecture Notes - Lecture 5: Rho Factor, Osmotic Concentration, Ribosomal Rna
eringoldfish322
BSCI 222
Final Exam
Study Guide
BSCI 222 Study Guide - Comprehensive Final Guide: Sister Chromatids, Wild Type, Ribozyme
mauveraccoon754

Related Questions

Place the following steps of DNA replication in right chronological order.
1. Primase adds an ribonucleotide primer
2. Initiator proteins bind the origin of replication and helicase breaks hydrogen bonds between complementary bases of double stranded template DNA
3. DNA polymerase III extends by complementary base pairing deoxyribonucleotides to the template strands
4. Ligase phosphodiester bonds the nucleotides to connect okazaki fragments nad connect fragments at the origin and where the "crotches or forks" collide
5. DNA polymerase I removes RNA primer and replaces with Deoxyribonucleotides

1) What is the difference between the leading strand and the lagging strand in DNA replication? Place the following steps of DNA replication in right chronological order.

1. Primase adds an ribonucleotide primer
2. Initiator proteins bind the origin of replication and helicase breaks hydrogen bonds between complementary bases of double stranded template DNA
3. DNA polymerase III extends by complementary base pairing deoxyribonucleotides to the template strands
4. Ligase phosphodiester bonds the nucleotides to connect okazaki fragments nad connect fragments at the origin and where the "crotches or forks" collide
5. DNA polymerase I removes RNA primer and replaces with Deoxyribonucleotides


2) Given the template DNA and primer sequences below, what bases will be added to the primer as DNA replication proceeds? (The bases should appear in the order that they will be added)

5'-GTCGTCTCCCTTTTTAAAAACCCCCCC-3'
5'-UUUUUAAAAAGGG-3'


3) Given the DNA sequence below, what could be a primer sequence used in its replication?
5'-CGTACGGCCCCAAAATTGCTACTCTGCTG-3'

1. In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that
DNA is a double helix.
DNA may be taken up by bacterial cells and be active.
DNA is present in bacteriophage T2.
eukaryotic cells may be genetically transformed.
DNA binds to the cell wall of the bacterium.
2. In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.
temperatures high enough to kill cells
enzymes that hydrolyze proteins
enzymes that hydrolyze DNA
enzymes that hydrolyze polysaccharides
enzymes that hydrolyze RNA
3. Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?
Pure RNA from S cells
Cell-membrane extracts from S cells
A mixture of pure RNA and protein from S cells
Cell-free extracts from S cells
Pure protein from S cells
4. A-T base pairs in a DNA double helix
form three hydrogen bonds with each other.
are more heat-stable than G-C base pairs.
are of a substantially different length than G-C base pairs.
are not accessible to DNA binding proteins.
are chemically distinct from G-C base pairs.
5. If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?
27
50
0
73
23
6. The uniform diameter of the DNA structure provides evidence for
antiparallel DNA strands.
a right-handed helix.
a double-stranded helix.
3′ end phosphorylation.
purine–pyrimidine pairing.
7. If a sequence of one strand of DNA is 5′-TGACTATC-3′, what is the complementary strand?
5′-ACTGATAG-3′
3′-TGACTATC-5′
3′-ACTGATAG-5′
3′-ACTGATAG-3′
5′-TGACTATC-3′
8. What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?
3′ end phosphorylation
Purine–pyrimidine pairing
Antiparallel DNA strands
Covalent bonds between sugars and phosphate groups
Hydrogen bonding between paired bases
9. If the Meselson–Stahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only 15N and the second only 14N, this result would have indicated that replication was
unidirectional.
dispersive.
conservative.
bidirectional.

semiconservative.
10. The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3′ hydroxyl group (–OH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.
The replication complex will no longer bind to origins of replication.
The DNA polymerase will only incorporate acyclovir molecules.
Acyclovir will bind single-strand binding proteins.
DNA polymerase will not be able to link a successive nucleotide.
Acyclovir will block the activity of the DNA helicase.
11. Which of the following does not demonstrate the stability of the DNA double helix?
Single-strand binding proteins are required to keep the strands apart.
Paired bases form hydrogen bonds.
High temperatures are required to denature it.
DNA synthesis requires energy.
DNA helicases require ATP.
12. What effect would a primase inhibitor have on DNA replication?
DNA polymerase would not be able to add bases to the DNA.
Primase would be blocked from joining the DNA replication complex.
Primase would not be able to provide primers for DNA polymerases.
DNA polymerase would not be able to use DNA as a template.
There would be no effect on DNA replication.
13. To replicate their DNA in a timely manner, most eukaryotic chromosomes
normally have uncoiled DNA.
have multiple origins of replication.
contain linear molecules of single-stranded DNA.
have two replication forks that move in the same direction.
are composed entirely of DNA.
14. Which statement about DNA replication is false?
Okazaki fragments are synthesized as parts of the leading strand.
Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
The sliding clamp increases the rate of DNA synthesis.
Replication forks represent areas of active DNA synthesis on the chromosomes.
Ligases and polymerases function in the vicinity of replication forks.
15. In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.
leading; telomerase
lagging; telomerase
lagging; DNA polymerase I
leading; DNA polymerase I
leading; replicase
16. A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?
DNA polymerase is constantly active.
Primase is able to extend the telomeres.
The mismatch repair system is extra efficient.
The telomerase gene is being expressed.
The cells do not need the ends of their chromosomes.
17. If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?
The replication complex excises the incorrect nucleotide.
The excision repair proteins remove the incorrect nucleotide.
The mismatch repair system scans for base-pairing mismatches.
There is no correction, because nucleotide errors by DNA polymerases are not repaired.
DNA ligase connects the correct nucleotide.
18. Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.
1, 2, 3, 4
2, 1, 3, 4
3, 4, 2, 1
2, 4, 1, 3
4, 2, 3, 1
19. Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.
64
32
12
128
6
20. When double-stranded DNA is heated to temperatures above 90°C, it denatures. Denaturation is a process that
breaks covalent bonds.
hydrolyzes DNA.
separates double-stranded DNA into two single strands.
causes new hydrogen bonds to form.
irreversibly destroys the double helical structure.