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Biotechnology
AP.BIO:
IST-1
(EU),
IST-1.P
(LO),
IST-1.P.1
(EK)
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Classroom
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The
flavonoid
3
-hydroxylase
(f3
'h)
gene
influences
flower
color
in
common
morning
glory
plants.
A
researcher
collected
DNA
from
common
morning
glory
plants
and
used
DNA
sequencing
to
identify
the
order
of
nucleotides
in
the
f3
'h
gene
for
each
plant.
A
subset
of
the
sequencing
data,
which
shows
the
results
for
two
plants,
is
shown
in
Figure
1.
Plant
1:
ATGGCTACCTTA
Position:
1
2
3
4
5
6
7
8
9
1011
12
Plant
2:
ATGCCTGCCTTA
1l
Position:
1
2
3
4
5
6
7
8
9
1011
12
Figure
1:
DNA
sequencing
data
for
positions
1-12
of
the
f3’h
gene
in
two
plants.
Which
of
the
following
can
be
determined
from
the
data
in
Figure
1?
Choose
1
answer:
°
CORRECT
(SELECTED)
Plant
1
and
Plant
2
differ
in
at
least
two nucleotides
of
the
f3'h
gene.
Figure
1
shows
a
type
of
DNA
sequencing
data
where
each
colored
peak
represents
the
detection
of
a
single
nucleotide
in
a
DNA
sequence.
According
to
the
data,
the
nucleotides
at
R
S
RS
I
S5
2N
TN
L7
N
SN
N
D
—_—
Y
SR
R
T
upe
R
W
]
.
Which
of
the
following
can
be
determined
from
the
data
in
Figure
1?
Choose
1
answer:
°
CORRECT
(SELECTED)
Plant
1
and
Plant
2
differ
in
at
least
two nucleotides
of
the
f3'h
gene.
Figure
1
shows
a
type
of
DNA
sequencing
data
where
each
colored
peak
represents
the
detection
of
a
single
nucleotide
in
a
DNA
sequence.
According
to
the
data,
the
nucleotides
at
positions
4
and
7
in
the
sequence
differ
between
the
plants:
Plant
1
has
a
G
and
an
A,
while
Plant
2
has
a
C
and
a
G.
So,
based
on
the
data
in
Figure
1,
we
can
determine
that
Plant
1
and
Plant
2
differ
in
at
least
two
nucleotides
of
the
f3’h
gene.
@
INCORRECT
Plant
1
does
not
produce
a
functional
F3'H
protein,
while
Plant
2
does.
Figure
1
shows
a
type
of
DNA
sequencing
data
where
each
colored
peak
represents
the
detection
of
a
single
nucleotide
in
a
DNA
sequence.
Using
this
data
we
can
see
that
Plant
1
and
Plant
2
have
nucleotide
differences
at
positions
4
and
7
of
the
f3’h
gene.
But,
we
need
more
information
about
the
sequence
of
the
gene
in
order
to
determine
how
this
nucleotide
difference
might
affect
protein
function.
@
INCORRECT
Plant
1
and
Plant
2
have
identical
genomes.
Figure
1
shows
a
type
of
DNA
sequencing
data
where
each
colored
peak
represents
the
detection
of
a
single
nucleotide
in
a
DNA
sequence.
Using
this
data,
we
can
see
that
Plant
1
and
Plant
2
have
nucleotide
differences
at
positions
4
and
7
of
the
f3’h
gene.
Because
of
these
differences,
we
know
that
Plant
1
and
Plant
2
do not
have
identical
genomes.
@
INCORRECT
Plant
1
makes
more
flowers
than
Plant
2
does.
Figure
1
shows
a
type
of
DNA
sequencing
data
where
each
colored
peak
represents
the
detection
of
a
single
nucleotide
in
a
DNA
sequence.
Using
this
data
we
can
see
that
Plant
1
and
Plant
2
have
nucleotide
differences
at
positions
4
and
7
of
the
f3’h
gene.
We
need
more
information
about
the
gene
to
predict
whether
this
change
might
impact
the
protein
function,
and
therefore
the
flower
phenotype.
Either
way,
this
difference
is
unlikely
to
affect
flower
number
because,
according
to
the
text,
the
f3’h
gene
influences
flower
color.
Way
to
go!
Review
related
articles/videos.
Report
a
problem
A
scientist
used
laboratory
techniques
to
move
Gene
X
from
a
bacteria-specific
plasmid
into
a
yeast-
specific
plasmid.
To
do
this,
the
scientist
first
used
restriction
enzymes
to
digest,
or
cut,
the
bacteria-
specific
plasmid
on
either
side
of
the
gene,
which
resulted
in
a
mixture
of
two
linear
pieces
of
DNA
(Figure
1).
Gene
X
Restriction
enzyme
digestion
fi
1,890
bp
I
)
686
bp
Figure
1.
Diagram
showing
the
bacterial
plasmid
before
and
after
restriction
enzyme
digestion
(bp
=
base
pairs).
The
scientist
used
gel
electrophoresis
to
separate
the
resulting
DNA
fragments
by
size.
Then,
the
scientist
cut
out
the
area
of
the
gel
containing
Gene
X
in
order
to
purify
the
gene
from
the
gel.
The
following
diagram
shows
the
digested
plasmid
DNA
as
visualized
by
gel
electrophoresis,
with
various
bands
labeled
A,
B,
and
C.
Which
of
the
bands
did
the
scientist
cut
out
in
order
to
isolate
Gene
X
DNA?
Choose
1
answer:
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1_12_2021
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10. The amino terminus of a wildtype enzyme in yeast has the amino acid sequence
Met-Leu-His-Tyr-Met-Gly-Asp-Tyr-Pro
A mutant, X, is found that contains an inactive form of the enzyme with the sequence
4
CO G Search or type URL
X
4 | 7 | - 125% + | @
Met-Gly-Asp-Tyr-Pro
at the amino terminus and a wildtype sequence at the carboxyl terminus. A second
mutant, Y, also lacks enzyme activity, but a full-length protein is not observed. Instead,
mutant Y makes a short peptide containing just 3 amino acids. What mutations could
account for these changes in mutant X and mutant Y? As part of your answer, show the
sequences of the genes encoding the wildtype proteins and the 2 mutant proteins, and
indicate in the wild type sequence where the mutations that gave rise to X and Y actually
occurred. What is the sequence of the…
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Select all that are true
1- traditional' (as nicknamed from the last 70-8- years) and organic farming are the same.
2- GOMs may involve including a gene from a completely unrelated organism, such as agene from a jelly fish into amonkey.
3- Traditional'( as nicknamed from the last 70-80 years) and GOM farming generally include the use of pesticides and herbicides
4-Organic food may not include genetics modifications, pesticides, or herbicides.
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€ 2 A
X
1_30*_SP23 - General Biology I (for majors)/11
f
us page
The anticodon sequence created from the following DNA: TACGGGGCTGAGATT
Select one:
a. AUGCCCCGACUCUAA
O b. Met-Pro-Arg-Leu-STOP
O c. UACGGGGCUGAGAUU
O d. Tyr-Gly-Ala-Glu-lle
20
000
MacBook Air
DII
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Materials
Updates
Grades
In some areas of the world, vitamin A deficiencies lead to a high number of
deaths every year. To combat this problem, scientists created golden rice, in
contrast to white rice. Using recmbinant technology, golden rice is created by
added two new beta-carotene genes, one from daffodil and one from a soil
bacterium. The result is a rice that contains high levels of vitamin A. In order to
Members
O Conferences
DBQ Online
create golden rice, which of the following processes is not employed?"
Newsela
OPolymerase chain reaction
nformation
P Blology Periods 1 and 2
Splicing
rading periods
ghschool MP1, Highschool
P2, Highschool MP3,
thschool MP4
Gel electrophoresis
O Gene therapy
cation
eting days
n Tue Wed Thu Fri
The sample furthest to the left contains the DNA size standards. What is the
INTL
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What is the mechanism of action for glyphosate?
(ii)
What are the implications for the long term use of Roundup herbicide? Suggest ONE
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while RNA is quantitatively degraded to 2'- and 3'-nucleoside
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Please help me?? immediately within an hour please????
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OPTIONS
a. Arg-Val-His
b. Lys-Val-His
c. Pro-Gly-Trp
d. Met-Leu-Ser
e.Thr-Ile-Phe
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10.18 Using the information in Figs. 10.14 and
10.18, explain how the two oligonucleotides
5'-CAAAGAAAAG-3' and 5'-CTTTTCTTTG-3'
assemble into a double helical structure (see Fig.
10.14 for the 3' and 5' numbering, and definitions
of C, A, G and T).
Fig. 10.18. Two strands of oligonucleotides sequenced
5-CAAAGAAAAG-3' and 5'-CTTTTCTTTG-3' assemble into a
double helix. The structure has been determined by X-ray diffraction
[M. L. Kopka et al. (1996) J. Mol. Biol., vol. 334, p. 653]. The
backbone of each oligonucleotide is depicted as an arrow pointing
towards the C3' end of the sequence, and the nucleobases are shown in
a 'ladder' representation. The nucleobases are colour coded: G, green;
A, red; C, purple; T, turquoise.
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2C & 2D
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A. =. A
Styles
Sensitivity
Font
Paragraph
Dictate
7. What are three differences between RNA and DNA?
7. Where is DNA found in the cell?
Where is RNA found in
the cell?
8. Name the three types of RNA. What is the function of each? Do they function in transcription,
translation, or both?
10. What are the steps of transcription?
11. What are the steps of translation?
12. If this is the base sequence of DNA, what is the resulting AA sequence for the following mutations,
where mutations and insertions are bolded and deletions are indicated with
DNA
TAC CGC T C C GCC G T C GA C A AT
АСС Аст
Mutations:
DNA
TAC CG C TC C GC C GT C GAC ACT
AC C A CT
DNA
TAC CGG T C C GC GT C GAC A aT
ACC AC T
DNA
TAC CG-
T C C GC GTC GACAAT
ACCAC T
DNA
TAC CG CA T CC GCC G T C GACA AT AC ACT
What are the consequences of each of these mutations for protein structure?
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Sensors detect the flash of
light.
DNA polymerase
Unused
deoxyribonucleotides are
cleaved by apyrase.
ATP is consumed by
luciferase and light is
emitted.
AMP and PP, are
converted into ATP by
sulfurylase.
Template strand
Growing strand
3' TAGGCCTACACTTACGCGAATGT 5'
5' ATCCGGAT 3'
dGTP
dNTPs
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dNMPs
+ P₁
PP₁
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Recommended textbooks for you
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Human Heredity: Principles and Issues (MindTap Co...
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