bios 390 molecular biology week 1 7 exam study guide with lab reports complete guide download to score an a 2021
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Eric Kyle
BIOS390 – Molecular Diagnostics
Week 1
(TCO 1) The basic principles of genetics contributed by Gregor Johann Mendel are:
1. The principle of segregation
2. The principle of independent assortment
(TCO 1) The Augustinian monk, Gregor Johann Mendel, crossed yellow-seeded and
green-seeded pea plants and then allowed the offspring to self-pollinate to produce
an F2 generation. The results were as follows: 6018 yellow and 2002 green (8020
total). The allele for green seeds has what relationship to the allele for yellow
seeds?
Yellow is dominant over the green seeds, where as green seeds is recessive.
(TCO 1) In 1953, James Watson and Francis Crick proposed
The double helix as a model for the structure of DNA.
Question Set 2
(TCO 1) In 1928, Frederick Griffith discovered that Streptococcus pneumonia
caused pneumonia in mice. In his experiments, mice were injected with different
strains of treated and untreated bacteria. Which of the following is not likely to have
occurred in the series of experiments by Griffith?
Live S bacteria inject into mice -die
Live R bacteria inject into mice-Live (do not have the capsule)
Heat killed S bacteria inject into mice-live
Mixture of live R bacteria and heat killed S bacteria injected into mice-Die
(TCO 1) In 1928, Frederick Griffith injected living S (smooth) Streptococcus
pneumonia into mice, and the mice died. When he injected living R (rough)
Streptococcus pneumonia into mice, the mice lived. When he injected heat-killed S
bacteria into mice, the mice lived. What was the result when he mixed heat-killed S
bacteria with live R bacteria and injected this mixture into mice?
Mixture of live R bacteria and heat killed S bacteria injected into mice – The
mice Die.
Griffith concluded that there was transformation which shows the role of DNA
as a genetic material.
,Eric Kyle
(TCO 1) In 1944, Oswald Avery, Colin MacLeod, and Maclyn McCarty used an in vitro
agglutination assay to demonstrate that purified DNA was sufficient to cause
transformation of cells, and that the transforming factor could be destroyed by
An enzyme called deoxyribonuclease that degrades DNA.
Question Set 3
(TCO 1) Beadle and Tatum’s “one gene-one enzyme” hypothesis was later revised
to the
“One gene – one polypeptide hypothesis”
(TCO 1) In the 1950s, Hershey and Chase conducted a now-famous experiment to
determine whether DNA or protein carried the hereditary information in
bacteriophage T2. What method did they used to selectively label the DNA and
protein components of bacteriophage T2?
Alfred Hershey and Martha Chase used radioactive sulfur and phosphorous to
trace the fates of protein and DNA, respectively, of T2 phages that infected
bacterial cells. They wanted to see which of these molecules entered the cells
and could reprogram them make more phages.
(TCO 1) The classic experiment performed by Alfred Hershey and Martha Chase
revealed
When the proteins were labeled (batch 1), radioactivity remained outside the
cells; but when the DNA was labeled (batch 2), radioactivity was found inside
the cells. Bacterial cells with radioactive phage DNA released new phages
with some radioactive phosphorus.
The researchers concluded that Phage DNA entered bacterial cells, but phage
proteins did not. Hershey and Chase concluded that DNA, not protein,
functions as the genetic material of phage T2.
The Hershey-Chase experiment was a landmark study because it provided
powerful evidence that nucleic acids, rather than proteins, are the hereditary
material, at least for viruses
Question Set 4
(TCO 2) The phosphodiester linkage between adjacent nucleotides to form a DNA or
RNA chain occurs by a _____ reaction involving the removal of _____ and
pyrophosphate.
,Eric Kyle
occurs by a condensation reaction involving the removal of H2O and
pyrophosphate
(TCO 2) The sugar in the DNA nucleotides is
a 5-carbon sugar (deoxyribose)
(TCO 2) What forms the “backbone” of a nucleic acid?
a chain of sugar and phosphate groups, linked through phosphodiester bonds
Question Set 5
(TCO 2) Which of the following did Watson and Crick know when they were trying to
determine the structure of DNA?
All except 6 and 8
applies…
(TCO 2) The predominant form of DNA in vivo is
B-DNA
, Eric Kyle
(TCO 2) Watson and Crick noted that DNA’s structure is interesting because it
suggested a possible copying mechanism. What about DNA’s structure facilitates
copying?
The strands of the double helix are complementary.
Question Set 6
(TCO 2) As DNA denatures, its absorption of UV light increases, a phenomenon
known as
the hyperchromic shift. The purine and pyrimidine bases in DNA strongly absorb
ultraviolet light.
(TCO 2) The melting temperature (Tm) of DNA is
The temperature at which the DNA strands are half denatured, meaning half double-
stranded, half single-stranded, is called the melting temperature(Tm). The amount
of strand separation, or melting, is measured by the absorbance of the DNA solution
at 260nm.
(TCO 2) When two strands of DNA from different sources are hybridized in the lab,
what provides the chemical stability for holding the two strands of DNA in a double
helix structure?
hydrophobic interactions (base-stacking) and hydrogen bonds
Question Set 7
(TCO 2) Hairpin loops, base-paired stems, and bulges make up what part of RNA’s
structure?
Base-paired stems in RNA form an A-type double helix.
A hairpin loop is an unpaired loop of messenger RNA (mRNA) that is created when
an mRNA strand folds and forms base pairs with another section of the same strand.
The resulting structure looks like a loop or a U-shape. A stem-loop structure is
formed in an RNA sequence when a short neighboring sequence consists of
complementary nucleotide sequence relative to its flanking regions. These
sequences have a high tendency to entangle and make a stem-loop like structure.
Importantly, the strength of this stem-loop structure depends upon the nature and
number of nucleotide bonds present. Generally, the more is the GC content of this
stem-loop binding, the more stable it is because GC bonds are comprised of triple
hydrogen bonds.
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