Bellringer
Chemical structures that are involved in
physiological processes, such as hemoglobin in
blood, insulin that regulates blood glucose
levels, and enzymes that regulate body
functions, are all made of proteins.
Name some parts of the human body that
contain proteins.
Construct a DNA model Using Base-pairing Rules.
Homework: 25 Pts. Due tomorrow to gain access to DNA Extraction Lab.
Template Strand
2 rings
G
A
T
T
A
C
A
C
T
G
T
C
A
G
A
A
A
C
C
T
A
A
T
G
T
G
A
C
A
G
T
C
T
T
T
G
3 bonds
1 ring
Directions:
Template Strand: _G_ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __
1) Answer Review Questions on back:
1. What is the basic subunit for DNA? Complementary Strand:___ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __
2. What are the three parts?
3. Which parts are the backbone?
4. Which nucleotides are purines?
Nucleotide Bank
5. Which nucleotides are pyrimidines?
6. How do you tell the difference between the two?
7. What type of bond holds together DNA strands?
2) Cut out the nucleotides from the Nucleotide Bank (right).
3) Match them up with the template strand above based
upon base-pairing rules then tape/glue beneath.
4) Determine the sequence (order of the string of
nucleotides read left to right) for both the template
strand and the new complimentary strand. Label the
sequence of the template and complimentary strand on
the lines with the nucleotide abbreviation.
Key Ideas
 What
is the process of gene expression?
 What
role does RNA play in gene expression?
 What
happens during transcription?
 How
do codons determine the sequence of amino
acids that results after translation?
 What are the major steps of translation?
 Do
traits result from the expression of a single gene?
Objectives: Transcription &
Translation
Today: Transcription
 Describe gene expression
 Explain the role of RNA in gene expression
 Summarize transcription
 Practice transcribing a gene.
Tomorrow: Translation
 Explain how codons determine the amino acid
sequence of a protein
 Describe the steps of translation
 Identify the complexity of gene expression
 Practice translating a gene.
Gene
Expression
How we get
from DNA to traits.

Vocabulary
 Gene
expression
 Transcription
 Translation
 RNA
 mRNA
 tRNA
 rRNA
 RNA polymerase
 Codon
An Overview of Gene
Expression
 So
far, we’ve discussed the structure of DNA…
being made of nucleotides that contain 1 of 4
different nitrogenous bases.
 You should know that DNA’s job is to store
genetic information.
 You’ve also learned the cell cycle.
 We’ll
be spending
the next few days on
G1 of the cell cycle.
 This is the phase
where most of
“living” takes place.
 It’s the phase that
proteins and traits
are made.
The Purpose of DNA…
 The
purpose of DNA itself is to house the
information necessary for heritable traits…meaning
that it holds the information from which proteins are
made.
 This is what living is all about.
 The DNA in our chromosomes is like books on the
shelf of a library… just waiting to be read.

DNA provides the original
information from which proteins
are made in a cell, but DNA
does not directly make proteins.
The Purpose of Life
An Overview of Gene
Expression,

Gene expression is the manifestation of genes
(contained in DNA) into specific traits.

What does manifestation mean?
THE CENTRAL DOGMA OF BIOLOGY =
DNA  mRNA  Protein  Trait
This process takes place in two main stages,
Transcription: the process of copying the
directions for traits out of DNA by making
mRNA
2. Translation: reading the directions copied in
RNA and turning them into the amino acid
sequences for the gene.



1.
THE CENTRAL DOGMA OF BIOLOGY
Directions to make HAIR COLOR
ACTGAACTGCACTG…
Genes:
The basic units of heredity.
They are located on specific regions of
chromosomes, contained in DNA.
There are thousands of genes “written” into each
of the 23 chromosomes in our cells.
They come from the original genes given to us
from mom & dad.



So What Does a Real Gene
Look Like?
Pro-melanin-concentrating hormone
tacagcgtgt ggcattctcc ccacattctc cttcggcttt acggagcagc
aaacaggatg gcgaagatga gcctctcttc ctacatgtta atgctggcct tttctttgtt
ttctcacggc attttacttt cggcctccaa gtccatcagg aacgtagaag cgacatagt
atttaataca ttcaggatgg ggaaagcctt tcagaaggaa ataccgcagaagatcggt
tgttgctcct tctctggaag gatacaaaaa tgatgagagc ggcttcatga
aggatgaaga tgacaagacc acaaaggtac gtgtatgcag tctgcctttt
attgcactag agatgaaaac gatgtttaca attataagcc acccagaagt
aaattttgta ttttaatttt ataaataggc tacatacag tcattgtgtg tattaagata
actaggaaaa cgtcatacaa accaggcatt tccccattct atccagaatc ttgtatcttg
tctcgcatat ggaggtaaag acagtataca gcatcttaga actgatcagc
aagaatgttg tacaactgta ttctagctct actctgaaga agacagctgg
gatacaaacc aatcttctct tcacagaaca caggctccaa gcagaatctc
gtaactcacg gtctgcccct cagtctggct gtaaaacctt acctcgctct
gaaaggacca gcagtcttcc cagctgagaa tggagttcag aatactgagt
ccacacagga aaagagggaa attggggatg aagaaaactc agctaaattt
cccataggaa ggagagattt tgacagtgag tagccttcta aacatgcaat
tcctacatat taattttata aaagagctct gagcttcact gagttggatc tgaccataac
aaaatcaaga ccatagttca gttctatcaa atagtaggca gcccacgtca
aaatggggaa tttttcaaaa tcagtaatag tggtttgttt tattctggat tcattataag
tccacagatt ctcttaattc tgtgtggtaa ttatagtcat tgtttgttcc ttttcagtgc
tcaggtgtat gctgggacga gtctaccgac cctgttggca agtctgatac
ctgctggtcc acaacatcct ttcagaagaa aacgattcat tgcaagtgga
gagaaaagcc cttaatgttg atgtaacttg tgtatcatcc taaatgtctg ttttaaaaga
aactggttac aatatgtaaa tgctatgtaa atgatatgct ttgacttgtg cattaaactt
cacaaaaatt ctgcata
-http://www.ncbi.nlm.nih.gov/gene/24659#reference-sequences
Hemoglobin Gene

taccacgacagaggacggctgttctggttgcagttccggcgga
ccccgttccaaccgcgcgtgcgaccgctcataccacgcctcc
gggacctctcctacaaggacaggaaggggtggtggttctggat
gaagggcgtgaagctggactcggtgccgagacgggtccaatt
cccggtccgttcttccaccggctgcgcgactggttgcggcacc
gcgtgcacctgctgtacgggttgcgcgacaggcgggactcgc
tggacgtgcgcgtgttcgaagcccacctgggccagttgaagtt
cgaggattcggtgacggacgaccactgggaccggcgggtgg
aggggcggctcaagtggggacgccacgtgcggagggacct
gttcaaggaccgaagacactcgtggcacgactggaggtttatg
gcaattcgacctcggagccatcgtcaaggaggacggtctacc
cggagggttgcccgggaggaggggaggaacgtggccggga
aggaccagaaacttatttcagactcacccgccg

http://www.bio.davidson.edu/courses/Bio111/Hemo
mut.html
Gene Transcription and Translation
Where Does it Occur?
RNA: A Major Player
 All
of the steps in gene expression
involve RNA.
 What
exactly is RNA, & how does is
compare to DNA?
 First,
like DNA, RNA is a nucleic acid
made of nucleotide subunits linked
together.
RNA vs. DNA


RNA is a nucleic acid like DNA
But RNA differs from DNA in 3 ways.
1.
First, RNA usually is composed of one strand
of nucleotides rather than two strands.
a. The exception occurs in viruses
2.
Second, RNA nucleotides contain the fivecarbon sugar ribose rather than the sugar
deoxyribose.
3.
Third, RNA nucleotides have a nitrogenous
base called uracil (U) instead of the base
thymine (T).
a. Uracil (U) is complementary to adenine (A)
whenever RNA pairs with another nucleic
acid.
DNA vs RNA Structure

Deoxyribose Nucleic Acid =
DNA
◦

Is missing one oxygen in the
ribose sugar.
Ribose Nucleic Acid = RNA
◦
Has all oxygens
Visual Concept: Ribonucleic Acid (RNA)
RNA: A Major Player
 In
cells, three types of RNA complement DNA and
translate the genetic code into proteins.
1. Messenger RNA (mRNA) is produced when
DNA is transcribed into RNA.
 The
mRNA carries instructions for making a
protein from a gene and delivers the
instructions to the site of translation.
RNA: A Major Player
2. Transfer RNA (tRNA) “reads” the instructions
carried by the mRNA at the site of
translation, then translates the mRNA
sequence into protein subunits called amino
acids.
3. Ribosomal RNA (rRNA) is an RNA molecule
that is part of the structure of ribosomes.
 Recall
from CH7, ribosomes are the cellular
structure where protein production occurs.
Objectives


Define Transcription
Summarize the steps of transcription

Step
1
2
3
In order to help keep this straight, make a chart like the one
below.
Major events
Transcription: Reading the Gene

Transcription is the process of creating a copy of a
gene in DNA as an mRNA molecule.

Transcription Steps
1.
INITIATION: Transcription begins when the enzyme
RNA polymerase binds to the specific DNA sequence
in the gene that is called the promoter.
-
The promoter’s role is to signal the RNA polymerase
where to start transcription.
The DNA always contains the sequence TAC for the
“start” signal.
Transcription: Reading the Gene
 Step
1
Transcription: Reading the Gene
2. ELONGATION: RNA polymerase then
unwinds and separates the two strands
of the DNA double helix to expose the
DNA bases on each strand. RNA
polymerase adds RNA nucleotides.
Transcription: Reading the Gene
 Step
1
Transcription: Reading the Gene,
3. TERMINATION: RNA polymerase moves along the
bases on the DNA strand and adds
complementary RNA nucleotides to a growing
mRNA as it “reads” the DNA of the gene until it
reaches the “stop” signal.

Remember that in transcription “U” matches with “A”, not
“T” like in replication.

-
-
-
The “A” still matched to “T” though.
As RNA polymerase moves down the DNA strand, a single
strand of mRNA grows.
Just as there is a “start” signal on the DNA, signaling the
start of the gene, there is a “stop” signal as well.
This region is specially designed to let the RNA polymerase
know when the gene ends & therefore when to stop
transcription.
This stop signal is one of 3 DNA sequences:
ATT, ATC, or ACT.
What would the RNA sequences be?
UAA, UAG, or UGA
Transcription: Reading the Gene
 Step
1
Visual Concept: Transcription
Transcription
Concept Check
 What
is the point of transcription?
 What enzyme is used in transcription?
 What are the signals for starting and
stopping a gene?
 Why is mRNA necessary?
Review










A gene is similar to a recipe.
Gene expression is like the process of baking a secret cake
recipe, complicated because the recipe is written in a language
the chefs don’t understand.
It is written as one long sentence composed of just one word.
The word is written in the language of the nitrogenous bases, A,
T, G, & C
A gene is written in a unique language that must be transcribed
by a messenger that speaks the language of the chefs.
The way the recipe is delivered to the chefs (the ribosomes) in
the cytoplasm (the bakery) is by the messenger “mRNA”.
mRNA copies the recipe during transcription and delivers it to the
bakery in the cytoplasm for translation to occur (decoding the
recipe in a different language to allow for baking the recipe).
In the cytoplasm the recipe is translated into the language of
proteins (amino acids) and finally made into proteins.
Finally, the secret cake is made.
Now it’s your turn to read the recipe.
Practice Transcription…making an mRNA
complement to the gene in DNA.
DNA= TCTACAGGAGCGCTGGCAAGACTGCCG
RNA= You make it.






Examine the DNA sequence above.
Look through and identify the promoter region containing the
“start signal” of DNA. Underline it.
Do the same for the “stop signal”.
Write an RNA sequence of bases using the complement to the
entire DNA sequence using the RNA bases (A-U-C-G), starting
with the sequence of the “start” site all the way until you reach
one of the 3 “stop” sequences.
Only write the RNA sequence that complements the DNA
sequence from the “start” to “stop” signals.
You have 5 minutes. Ask questions if you need to.
DNA: Practice…
TCTACAGGTGCAAGACTGCCG
mRNA:
o Find the start sequence…underline it.
o Find the stop sequence…underline it.
o Starting with the start sequence, transcribe the gene using the
RNA bases.
o What you end up with is an mRNA transcript of the gene
contained in the DNA.
In-class Exercise/HW


Practice Transcribing:
You’re going to play the role of the messenger now. You need to
be able to take a DNA sequence and identify the mRNA that will
“copy” the recipe, the gene for a protein, so the recipe can be
made by the ribosomes.
Gene Xlr23:
CGAACCTACAGTTCCGCGTCGGGCTAGACTGGCAATG
Copy this sequence down on a sheet of paper.
2. Identify the “start” sequence within the DNA above
(underline it).
3. From the start sequence, count in groups of three until
you reach one of the three “stop” signals.
4. What is the “stop” sequence (underline it).
5. Just below the DNA sequence you copied, transcribe
the DNA into a sequence of mRNA for the gene Xlr23.
Tomorrow we will use this sequence to practice translation.
1.
Reflections
 What
did you learn today?
 Design
an acronym of pneumonic device to
remember the types of RNA and steps in
transcription.
In-class Exercise/HW


Practice Transcribing:
You’re going to play the role of the messenger now. You need to
be able to take a DNA sequence and identify the mRNA that will
“copy” the recipe, the gene for a protein, so the recipe can be
made by the ribosomes.
Gene Xlr23:
CGAACCTACAGTTCCGCGTCGGGCTAGACTGGCAATG
Copy this sequence down on a sheet of paper.
2. Identify the “start” sequence within the DNA above
(underline it).
3. From the start sequence, count in groups of three until
you reach one of the three “stop” signals.
4. What is the “stop” sequence (underline it).
5. Just below the DNA sequence you copied, transcribe
the DNA into a sequence of mRNA for the gene Xlr23.
Tomorrow we will use this sequence to practice translation.
1.
Objectives Day 2
 Explain
how codons determine the amino
acid sequence of a protein
 Describe the steps of translation
 Identify a complexity of gene expression
 This
is a short lecture so stay focused.
The Genetic Code: ThreeLetter “Words”
What is the mRNA you decoded for gene Xlr23?
DNA C G A A C C T A C A G T T C C G C G T C G G G C T A G A C T G G
AUGUCAAGGCGCAGCCCGAUCUGA
mRNA

◦


What do you notice that is similar about the start and
stop sequences?
There is significance in the number 3 in RNA.
◦

Save this and we’ll move on…
It corresponds to what’s called a codon.
A codon is a three-nucleotide sequence in mRNA.
The Genetic Code: ThreeLetter “Words”
A
codon is a key that corresponds to 1 of 20
amino acids.
 An
amino acid is the building block of a protein.
 Codons
also act as the start or stop signal for
translation.
 These
“signals” are referred to as “start” and “stop”
codons on mRNA in genetics.
 So
the start codon is…AUG (signals the start
of the gene)
 The stop codons are… UAA, UGA, UAG
(signals the end)
The Genetic Code: ThreeLetter “Words”
 Refer
to you handout.
 There are 64 mRNA codons.
 The
mRNA that is created in transcription
is actually a collection of a series of 3nucleotide sequences called codons.

So each gene will contain nucleotides in
multiples of 3
The Genetic Code: ThreeLetter “Words”
Your practice from last night…
AUGUCAAGGCGCAGCCCGAUCUGA
AUG-UCA-AGG-CGC-AGC-CCG-AUC-UGA
 Notice that the length of the gene is a
multiple of 3…
 This is the way all genes are…in
multiples of 3.
 This is why I asked you to count by
threes until you reached the “stop”
codon.

The Genetic Code: Three-Letter
“Words”
 Each
codon specifies for only one amino
acid, but several amino acids have more
than one codon.
 See leucine
 This system of matching codons and amino
acids is called the genetic code.
 The genetic code is based on codons that
each represent a specific amino acid.
 This is the translation tool that helps to
translate the mRNA from the nucleotide
language into the language of amino
acids.
Figure 13. The amino acid coded be a specific mRNA codon can be determined by
following the three steps below. What amino acid does the codon GAA code for?
Codons in mRNA
Translation: RNA to Proteins
 Translation
is the process that changes the
mRNA molecule into the complementary
amino acid sequence.
 Takes place in the cytoplasm
 occurs in a sequence of 5 steps
 Involves all three kinds of RNA
 and results in a complete polypeptide.
Translation: RNA to Proteins
 Translation
relies upon the tRNA
molecule to act as the gobetween for mRNA codon &
the amino acid that
corresponds to it.
 There
is only one specific amino
acid for each codon.
 The mRNA gets matched up
with the right tRNA molecule
because of the anti-codon
region
 An
anticodon is a threenucleotide sequence on tRNA
that is complementary to an
mRNA codon.
 There
are two
important
regions of a
tRNA.
 The area where
the amino acid
attaches &
 The anticodon
region, which is
complementary
to the codon of
mRNA
 The anticodon
always decides
which amino
acid is carried.
AMINO ACID
GOES HERE
ANTICODON
tRNA matches
mRNA here
The Steps to Translation
Translation: RNA to Proteins,
 Step 1
 A ribosome
attaches to the mRNA
 The
UAC (methionine) tRNA attaches to the
start codon on mRNA within the ribosome.
 Step 2
 The tRNA
molecule that has the correct
anticodon and amino acid binds to the second
codon on the mRNA.

A peptide bond then forms between the two amino
acids, and the first tRNA is released from the ribosome.
Translation: RNA to Proteins,
 Step
3
 The ribosome then moves one codon down
the mRNA, kicking the 1st tRNA out.
 The
amino acid chain continues to grow as
each new amino acid binds to the chain and
the previous tRNA is released.
 Step 4
 This process is repeated until one of three stop
codons is reached.
 A stop codon does not have an anticodon, so
protein production stops.
Translation: RNA to Proteins,


Step 5
The newly made polypeptide falls of the ribosome, the
tRNA leaves the ribosome, & the ribosome falls apart.

Translation is complete & the polypeptide is free to go
get processed into a protein in either the ER or the
Golgi.

This is where translation ends but it doesn’t have to be
the only protein made.
Repeating Translation


Many copies of the same protein can be made rapidly
from a single mRNA molecule because several ribosomes
can translate the same mRNA at the same time.
Translation: RNA to Proteins
PEPTIDE BOND FORMS:
Then the ribosome moves forward
Methionine
RIBOSOME
Anticodon
UAC
mRNA
AUG UCA AGG CGC AGC CCG AUC UGA
Start Codon
Other amino codons
Stop Codon
Methionine Serine
Anticodon
UAC AGU
mRNA
AUG UCA AGG CGC AGC CCG AUC UGA
PEPTIDE BOND FORMS:
Then the ribosome moves forward
Methionine Serine
The growing chain of
amino acids is a
polypeptide,
or in other words…a
protein
Anticodon
mRNA
AGU
AUG UCA AGG CGC AGC CCG AUC UGA
PEPTIDE BOND FORMS:
Then the ribosome moves forward
Methionine Serine
Anticodon
mRNA
?
UCC
AUG UCA AGG CGC AGC CCG AUC UGA
PEPTIDE BOND FORMS:
Then the ribosome moves forward
Methionine Serine
Anticodon
mRNA
?
?
GCG
AUG UCA AGG CGC AGC CCG AUC UGA
PEPTIDE BOND FORMS:
Then the ribosome moves forward
Methionine Serine
Anticodon
mRNA
?
?
?
UCG
AUG UCA AGG CGC AGC CCG AUC UGA
PEPTIDE BOND FORMS:
Then the ribosome moves forward
Methionine Serine
Anticodon
mRNA
?
?
?
?
GGC
AUG UCA AGG CGC AGC CCG AUC UGA
Methionine Serine
Anticodon
mRNA
?
?
?
?
?
UAG
AUG UCA AGG CGC AGC CCG AUC UGA
Once the stop codon is reached
translation terminates. There is
no tRNA for the stop codon so
the ribosome know to detach.
The newly formed polypeptide
then leaves to get processed.
Complexities of Gene Expression



The relationship between genes and their effects is
complex.
Not 1 simple outcome…



1 gene = multiple traits
Multiple genes required for 1 trait
1 gene = 1 trait

Some genes are expressed only at certain times or
under specific conditions.
Variations and mistakes can occur at each of the
steps in replication and expression.
The final outcome of gene expression is affected by
the environment of the cells, the presence of other
cells, and the timing of gene expression.
The environment can also affect gene expression.



In summary, one gene can be used for many trait
outcomes.
7 different trait possibilities for the same gene.
Summary

Gene expression produces proteins by
transcription and translation. This process
takes place in two stages, both of which
involve RNA.

In cells, three types of RNA complement DNA
and translate the genetic code into proteins.

During transcription, the information in a
specific region of DNA (a gene) is transcribed,
or copied, into mRNA.
Summary, continued

The genetic code is based on codons that
each represent a specific amino acid.

Translation occurs in a sequence of steps,
involves three kinds of RNA, and results in a
complete polypeptide.

The relationship between genes and their
effects is complex. Despite the neatness of
the genetic code, every gene cannot be
simply linked to a single outcome.
In Class Exercise
CGAACCTACAGTTCCGCGTCGGGCTAGACTGGCAATG
AUGUCAAGGCGCAGCCCGAUCUGA
 Complete the Gene.
 Translation is the last step of gene expression as it forms
the final polypeptide. Your exercise today is to take the
gene we transcribed into mRNA yesterday and
translate it into a polypeptide. Write your polypeptide
as a series of circles with the name of the
corresponding amino acid within. This is the protein for
the gene you transcribed.
Methionine


?
?
?
?
?
Check these off with me to make sure you got it…
 Tomorrow
?
you’ll have to transcribe & translate a much
bigger gene.
AUG UCA AGG CGC AGC CCG AUC UGA
Codons in mRNA
 mRNA
How are you progressing?
Answers to IC/HW Exercise
=
AUGUCAAGGCGCAGCCCGAUCUGA
 Poly peptide chain =
Isoleucine
Practice the Process: Find, transcribe and
translate the gene into a polypeptide sequence.
 DNA:
 mRNA:
GCAATACGTAAATAGATCTATCGC
 AUG
 Polypeptide:
CAU UUA UCU AGA UAG
 Met-His-Leu-Ser-Arg-(stop)
“Rosetta Stone” of Genetics
Complement
Gene
(DNA)
mRNA
(codon)
Anticodon
T
A
U
A
A
T
A
U
G
C
G
C
C
G
C
G
Fill in the lines for the following
sequences.

ATGACTAGCTGGGGGTATTACTTTTAG
Complement:___________________________________________________________________

Gene:
TACTGATCGACCCCCATAATGAAAATC
____________________________________________________________________

mRNA:
AUG - ACU - AGC - UGG - GGG - UAU - UAC - UUU - UAG

tRNA:
UACUGAUCGACCCCCAUAAUGAAAAUC
___________________________________________________________________

AA:
MET-THR-SER-TYR-GLY-TYR-TYR-PHE-STOP
___________________________________________________________________
Fill in the lines for the following sequences.
Complement
ATG
Gene
TAC
ATG
mRNA
tRNA
Amino Acid
UGU GAU
CUC
UUG
ala
AUU
pro
Fill in the lines for the following sequences.
Complement
ATG
GAG TGT
GAT
GCC TAC
AAC
CCT
Gene
TAC
CTC
CTA
CGG ATG
TTG
GGA ATT
mRNA
AUG
GAG UGU
GAU
GCU
UAC
AAC
CCU
tRNA
UAC
CUC
ACA
CUA
CG
AUG
UUG
GGA AUU
Amino Acid
MET
GLU
CYS
AST
ala
TYR
ASP
pro
XXX
ACA
asparta
te
GCC
GCA
GCG
TGTGAT(GCC…)TACAAC(AAC…)TAA
XXXCTCACACTA(CGG…)XXXTTG(GGA…)ATT
AUGGAGXXXXXX(GCC…)UACAAC(CCU…)UAA
UACXXXACACUA(CGG…)AUGXXX(GGA…)XXX
MET-GLU-CYS-VAL-XXX-TYR-THR-XXX-STOP
asparagi
ne
CCC
CCA
CCG
TAA
UAA
TRX/TRL: CW/HW: Using the Genetic Code HW
Genetic Code of Keratin
Keratin is one of the proteins in hair. The gene for keratin is transcribed and translated by certain
skin cells just underneath the growing hair. The sequence below is part of the mRNA molecule that
is transcribed from the gene for keratin.
Analysis
1. Determine the sequence of amino acids that will result from the translation of the segment of
mRNA above. Use the genetic code in Figure 13.
_____Methionine – Serine – Arginine – Glutamic Acid – Phenylalanine – Serine - __________
2. Determine the anticodon of each tRNA molecule that will bind to this mRNA segment.
_____UAC – AGA – GCA – CUU – AAA – AGG
______________________________________________________________________
3. Critical Thinking Recognizing Patterns: Determine the sequence of nucleotides in the segment
of template DNA from which this mRNA strand was transcribed.
____TAC – AGA – GCA – CTT – AAA – AGG
______________________________________________________________________________
4. Critical Thinking Recognizing Patterns: Determine the sequence of nucleotides in the segment
of DNA that is complementary to the DNA segment that is described in item 3.
____ATG – TCT – CGT – GAA – TTT – TCC
______________________________________________________________________________
Transcription/Translation Lab
 Working
with a partner, analyze the gene
Xlr24 to determine the DNA sequence of
the gene, the sequence of the mRNA,
and the sequence of the amino acids
that will form the polypeptide.
 Answer the associated questions.
 This is a 100pt lab due next Monday.