Nucleic Acids Convey Information: A View Of Gene Expression

I. Introduction

A. Introduction: Molecules Are Necessary For Life1. -“Life…..is a relationship between molecules”Linus Pauling

2. Important molecules studied so far

a. DNAb. RNAc. Protein

3. The idea that molecules could provide important functions like carry information began in the 1800s with the work of Friedrich Meischner

4. By the middle part of the 20th century it became known that DNA carries genetic information, thus confirming that genetic information is carried in molecules

a. Avery, McCleod and McCartyb. Hershey and Chase

5. By the middle part of the 20th century biochemists had readily showed that enzymes catalyzed many important reactions within cells

6. Work by Beadle and Tatum showed that genes work by controlling the synthesis of enzymes (one gene-one enzyme hypothesis)-all of which are proteins

7. Question: How do genes direct synthesis of proteins?

B. Introduction: Molecules Are Necessary For Life

“Just as our present knowledge and practice of medicine relies on a sophisticated knowledge of human anatomy, physiology and biochemistry, so will dealing with disease in the future demand a detailed understanding of the molecular anatomy, physiology and biochemistry of the human genome…..We shall need a more detailed knowledge of how human genes are organized and how they function and are regulated. We shall also have to have physicians who are as conversant with the molecular anatomy and

physiology of chromosomes and genes as the cardiac surgeon is with the structure and workings of the heart. “Paul Berg – Nobel Laureate 1980

C. Introduction: Understanding How Molecules Affect Human HealthUnderstanding molecular functions is important to modern medicine

1. Being able to understand gene expression helps with understanding of how to treat or prevent single gene disorders

2. Single gene disorders arise from a mutation(s) in a single gene which will lead to disease

a. Cystic fibrosisb. Muscular dystrophyc. Tay-Sachs Disease d. Sickle Cell Anemia

3. Being able to understand gene expression helps in the study of more complex diseases

a. Cancerb. Alzheimer’s diseasec. Diabetes

4. Being able to understand gene expression allows for the development of treatments and vaccines to pathogens

a. As early as the 1940s, Molecular Biologists knew that viruses were made of protein and nucleic acid

b. By 1952 Hershey and Chase showed that viruses held their genetic information in nucleic acid

c. Today, the study of which viral genes are being expressed at important parts of their life cycle leads to the development of treatments and vaccines

D. Introduction: Understanding Gene Expression

1. From the point where it was known that DNA was the prime genetic molecule, how it worked became a key point of study

2. Two key questions arose when it came to the study of DNAa. How genes are expressed (How does information in DNA encode proteins)

b. How is the DNA replicated

3. For each process, there are a series of proteins involved in mediating the process

a. Enzymesb. Proteins that become part of larger structures

II. The Central Dogma

A. The Central Dogma: Introduction

1. The flow of information involving the genetic material is termed as the “central dogma,” which was coined by Francis Crick

2. He coined this phrase due to the fact that “once information is passed into protein it cannot get out again”

3. The central dogma is basically a shorthand way of saying how genes get expressed: DNA mRNA Protein

4. The protein carries out the function of the gene

5. The central dogma has held up, but has been modified to include genes whose final product is an RNA

a. rRNAb. tRNAc. miRNAd. siRNAe. snoRNAf. snRNA

B. The Central Dogma: The Flow of Genetic Information

1. The flow of genetic information is indicated by the arrows in the classic central dogma figure

2. The arrow that flows from DNA to RNA indicates a process known as transcription

a. DNA is used as a template to produce RNA

b. If the RNA produced is a messenger RNA, then this is not the terminal step in gene expressionc. For all other RNAs, this is the terminal step

3. The arrow that flows from RNA to protein indicates a process known as translation

a. mRNA is the template for translationb. No other RNAs serve as templates

4. The arrows that flow from DNA to RNA, as well RNA to Protein are in most cases unidirectional

a. Never a flow from protein back to RNAb. Can be a flow of information from RNA back to DNA by a process known as reverse transcription

5. The round arrow around DNA indicates that DNA can serve as a template for its own replication

III. Basic Mechanisms of Gene Expression

A. Basic Mechanisms of Gene Expression: The Structure Of A Gene

1. In order to understand how a gene gets expressed, we must look at the basic structure of a gene

2. In order to look at the gene, we must understand which are the coding and non-coding strands

3. When we draw out our DNA molecule, the top strand is usually written in the 5’ 3’ direction left to right-this is considered the coding strand because this strand will have the same sequence as the mRNA that will be transcribed (except for U instead of T)

4. When we draw our DNA molecule, the bottom strand is the non-coding strand-this is the template for transcription

5. Each protein coding gene has two basic units

a. The transcriptional unit which are the sequences which will be transcribed into mRNAb. Upstream (5’) to the transcriptional unit is the promoter which directs transcription

6. When we number the base pairs in a gene, we give the base pair where transcription starts, the number 1 (based off of the coding strand)

a. Any base pair 3’ to the transcriptional start will have a positive number (going away from the transcription start site)b. Any base pair 5’to this transcriptional start will have a negative number (going consecutively away from the transcription start site)

7. The promoter is generally close to the actual gene itself and is found between -100 bp and about -35 bp

B. Basic Mechanisms of Gene Expression: Transcription

1. Transcription is the first step in gene expression

2. The main goal of the process of transcription is to produce a pre-mRNA

3. When a pre- mRNA is transcribed, it is transcribed by the RNA Polymerase II holoenzyme in the 5’ 3’ direction using

the noncoding strand as a template

4. As RNA polymerase II transcribes the pre-mRNA it will add nucleotides that have a complementary nitrogenous base to that of the template (therefore, the pre-mRNA has complementary sequence to the non-coding strand of DNA)

5. The pre-mRNA is essentially single stranded copy of the coding strand of the DNA

a. With RNA, ribose is used instead of deoxyriboseb. Uracil is used instead of Thymine

6. Figure 2-12

C. Basic Mechanisms of Gene Expression: Pre-mRNA Processesing

1. As we talked about, a pre-mRNA is produced by transcription

2. This is not the mature form of mRNA that is used in later steps of gene expression

3. In order to get the mature mRNA, the pre-mRNA must undergo some processing

4. The pre-mRNA contains two types of sequences

a. Exonsb. Introns

5. Exons are the portions of the pre-mRNA that remain and will compose the mature mRNA

6. Introns are considered intervening sequences, and are removed from the pre-mRNA by a process known as RNA splicing

7. For many genes, they will encode pre-mRNAs with more sequence devoted to introns than exons

8. For example, the dystrophin gene on the X-chromosome spans more than 2 million base pairs, and less than 1% of those base pairs consists of exon sequence

9. Splicing is not the only event that occurs in mRNA processing

10. In mRNA processing a 5’ cap is placed on the mRNA

11. In mRNA processing, there should be a polyadenylation signal (5’ AAUAAA 3’) towards the 3’ end. At this signal, the pre-mRNA is cleaved and a poly-A tail is placed on the mRNA

12. In the end, the mature mRNA is composed of the exons, a 5’ cap and a poly-A tail on the 3’ end

13. The 5’ cap and the poly-A tail are important in protecting the mature mRNA from degradation, and for stimulating translation of the mRNA

14. Once the mRNA is mature it is transported to the cytoplasm for the next steps in gene expression

D. Mechanisms of Gene Expression: Ribosomes Form A Structure Which Translates the Information in An mRNA

1. Once a mature mRNA arrives in the cytoplasm it immediately becomes available for translation

2. Translation is carried out by structure known as the ribosome

3. The ribosome is made of two subunits which come together on the mature mRNA to translate the mRNA

a. 40S subunit (in Eukaryotes)b. 60S subunit (in Eukaryotes)

4. Each subunit consists of proteins and rRNA

5. The role of the ribosome is to translate the mRNA sequence three bases at a time into a single amino acid, and a protein is made up of a polymer of amino acids

E. Basic Mechanisms of Gene Expression: The Discovery of mRNA

1. The first type of RNA found in the cell was rRNA

a. Most abundant type of RNA in the cell (80-85%)b. Been shown to associate with ribosomal proteins and become stably part of the superstructure of the ribosome

2. The discovery of mRNA owes itself to the study of the T4 bacteriophage (1960)

a. Virus that infects E. colib. Upon infection, E. coli cells stop synthesizing their own RNAc. The only type of RNA produced is T4 RNA

3. After transcription, the T4 RNAs became associated with ribosomes, but it does so in a different way than the E. coli rRNA

4. Instead of being incorporated into the super structure, the T4 RNA move across the surface

5. This moves the nitrogenous bases of T4 RNA into positions to be translated

6. Soon after the discovery of how T4 RNA worked, a class of E. coli RNA was discovered that worked in the exact same way, and was called mRNA (messenger RNA)

F. Basic Mechanisms of Gene Expression: The Structure of a Mature mRNA

1. The DNA sequence is read in triplets (3 bases) called codons through an mRNA intermediate

a. The mRNA will have the same nitrogenous base sequence of the coding strand

b. All thymines will be replaced by uracils in the mRNA

2. An mRNA has three basic units

a. 5’UTR (Untranslated Region)b. Open Reading Frame (ORF) c. 3’ UTR (Untranslated Region)

3. The 5’UTR serves as a site of mRNA regulation, also it is involved in the initiation (start) of translation

4. The open reading frame is the part of the mRNA that codes for a protein

a. Also known as the protein coding regionb. Begins with the presence of a start codon on the 5’ end and ends with a stop codon on the 3’ end

5. The typical start codon is an AUG and encodes the amino acid methionine

a. AUG, methionine is the start codon for 99% of all proteinsb. GUG (valine) & CUG (leucine) are also used as start codons, but are rare

6. There are three possibilities for stop codons

a. UAA, UGA and UAGb. These codons do not encode an amino acid, but instead tell the ribosome to stop translating

7. The 3’UTR serves as a site of mRNA regulation

G. Basic Mechanisms of Gene Expression: Translation

1. The goal of the process of translation is to translate the genetic code into a sequence of amino acids that compose a protein

2. In order to start translation, the 40S subunit binds the 5’ cap of the mRNA and scans down the mRNA for the start codon (AUG)

3. Once the 40S subunit finds the start codon, it stalls and waits for the 60S subunit to join

4. Once the 60S subunit joins, translation commences

5. There are a possible 20 amino acids that can be incorporated into proteins, but only 4 nitrogenous bases found in an mRNA

6. This led to the idea that multiple nitrogenous bases must encode for a single amino acid

7. There are only 16 possible two base codes and 20 total amino acids, so therefore more than two bases must encode a single amino acid

8. There are 64 possible three base codes and a total of 20 amino acids, therefore three base codes are necessary for encoding a single amino acid

a. Each 3 base code is called a codonb. Each amino acid can be encoded by multiple codons

9. To figure out what amino acid each codon encoded, synthetic RNAs were used

10. Let’s take an RNA that has the sequence 5-P-CCC UAG GCA CUC AUG CCC UUU GCU GGG UAA-OH 3’ and translate it

11. The start codon as we know is AUG

12. Therefore, we must find the first AUG in our mRNA

13. The first AUG is located starting at nucleotide 13. Since we see an AUG, we start translation with a MET (by using the chart on the right) This methionine is located at the amino terminus

14. The start codon establishes the reading frame for the mRNA

a. Each mRNA will have 3 sets of 3 base codes which can be readb. In other words the placement of the start codon establishes the remaining codons by moving 3 bases each time towards the 3’end of the mRNA

15. Each “codon box” is composed of four three-letter codes

a. 64 possible three letter codesb. All but three encode amino acidsc. The codon table is read with the first base in the codon on the left, the second base in the codon on top and the third base in the codon being read on the right

16. The next codon is CCC. By using the chart on the right is Proline

17. The next codon is UUU, which encodes for a Phenylalanine

18. Therefore, we have a protein with the sequence of N terminus-MET-PRO-PHE-ALA-GLY-C terminus

19. Note, the glycine is located at the carboxy (C) terminus

20. Within the code, there are three codons that do not encode amino acids, instead they tell the ribosome to stop

a. They are called stop codonsb. Premature stop codons can lead to defects in gene expression, and may lead to disease

21. Synthesis of the new peptide occurs from the amino terminal-to the carboxy terminal, in much the same way the mRNA is read in the 5’ 3’ direction

a. Where the peptide chain begins is known as the amino terminalb. Amino acids are added onto the carboxy terminal

H. Basic Mechanisms of Gene Expression: Translating the Genetic Code

1. The genetic code, which was deciphered about 50 years ago provides the fundamental clues for decoding the information in the mRNA into polypeptides

2. The genetic code is said to be degenerate

a. In most cases, it is the identity of the third base that does not matter when specifying which amino acid a codon will encode

b. Ex. Leucine is encode by four different codons (CU_)

3. This obseration gave rise to the “wobble hypothesis”

a. Proposed by Francis Crickb. Pairing between codon and anticodon at the first two codon positions always follows the usual rule of complementary base pairingc. Exceptional “wobbles” (non-Watson-Crick base pairing) can occur at the third position.

J. Mechanisms of Gene Expression: The Genetic Code Is Not Exactly The Same For All Organisms

1. Initially it was thought that the genetic code was universal for all organisms

2. For some organisms and organelles some codons encode different amino acids than what appears on the standard chart

a. CUG in Candida albicans encodes serineb. UGA encodes tryptophan in mitochondria

3. Sometimes UGA can encode selenocysteine

a. The 21 amino acid b. Is essential for mammalian development as knocking out the tRNAsel gene results in emryonic lethality (in mice)c. Found in greater than 40 genes that code for antioxidants and the type I iodothyronine deiodinase of the thyroidd. Yeast and plants do not appear to possess the machinery to insert selenocysteine into proteins

4. UAG can sometimes encode pyrrolysine

a. The 22 amino acidb. Found in archaea and eubacteriac. Found in some methylamine methyltransferases

K. Mechanisms of Gene Expression: Translation and the Adaptor Hypothesis

1. Once Molecular Biologists were able to effectively translate the genetic code (determine which amino acid is encoded by each

codon), the focus of study turned to the mechanisms of how the genetic code was translated

2. At first, it was thought that the mRNA would create cavities on their outer surfaces

a. Each codon would create a cavity of slightly different shapeb. The cavity shape would specify the appropriate amino acid

3. Francis Crick did not believe this hypothesis to be correct due to the chemistry of RNA and of the various amino acids

a. Certain groups on the nitrogenous bases are able to form hydrogen bonds, and thus would be more likely to interact with only polar or charged side chains of amino acids

b. Most amino acids have non-polar side chains making it difficult for the nitrogenous bases to interact with them

c. It would be unlikely that the folded RNA would be able to discriminate between amino acids with similar chemical structure

4. This led Francis Crick to believe that there would be an adaptor molecule that would bring the correct amino acid to the ribosome during translation

5. Crick postulated that an RNA molecule may serve as an adaptor because it would be able to base pair to the codon in the mRNA, and thus appropriately “read” the information in the mRNA

6. Paul Zamecnik and Mahlon B. Hoagland used cell free extracts to perform experiments to study how translation worked

a. Discovered that transfer RNAs (tRNAs) serve as adaptor molecules to bring the correct amino acid to the ribosome during translationb. Each individual tRNA has an anticodon, which base pairs to the appropriate codonc. The anti-codon present in the tRNA will specifiy the amino acid that will be attached to it

d. Each tRNA with a different anti-codon is encoded by a different gene

L. Mechanisms of Gene Expression: The Role of tRNA In Translation

1. In actual translation, the tRNA (transfer RNA) brings the appropriate amino acid to the ribosome

2. Each individual tRNA will have its own anti-codon, and will be charged with the appropriate amino-acid

3. Bonding between the anti-codon and the codon brings the appropriate amino acid into the next position for adding on to the growing peptide chain

4. For the stop codons, there is no corresponding tRNA

5. When a stop codon is reached, the peptide is completed and released from the ribosome

IV. Protein Structure and Function

A. Protein Function: Introduction

1. Once the mRNA is translated, we have protein-yet the process of gene expression is not complete

2. In order to have a protein with function, it must fold properly into its 3-dimensional structure

3. Each protein will have a specific folded structure that will correlate nicely to its function

4. Proteins have a variety of diverse functions in cells

a. Act as structural components to give cells shapeb. Hormonesc. Enzymes

5. As the protein is being synthesized during translation, it is folding into its optimal 3D structure, such that it functions properly in the cell

a. Occurs co-translationallyb. Peptide is still associated with the ribosome

6. The folded structure of the protein are dependent on two factors

a. The intrinsic sequence of amino acids within the protein (primary structure) b. External influences known as molecular chaperones

7. The role of the molecular chaperone is to aid the newly synthesized protein fold into its proper conformation for

8. Most proteins require molecular chaperones to fold properly in vivo

9. Molecular chaperones are proteins that act to properly fold newly produced proteins

B. Protein Structure and Protein Folding: Introduction

1. Overall, there are four levels of proteins structure

a. Primaryb. Secondary c. Tertiaryd. Quartenary

2. The ability for a protein to fold through these levels lies initially in the chemistry of the amino acids that compose the protein

3. Structurally, each amino acid has these groups attached to a central carbon ( carbon)α

a. An amino group (NH3+)

b. A carboxyl group (COO-) c. A hydrogen d. An R group – this R groups is variable amongst the amino acids and gives each amino acid its specific identity e. Note: each R group will have its own identity; some may be charged, polar or hydrophobic

4. At pH=7, the amino and carboxyl groups are charged, but over a pH range of 1-14, these groups exhibit binding and dissociation of a proton

5. The amino and carbonyl groups play intricate roles in forming secondary structure, whereas the R-groups play important roles in the formation of more intricate protein structures

C. Protein Structure: Primary Structure

1. To produce the primary structure of a protein, amino acids must be joined together by a specific bond known as a peptide bond (Job of the ribosome)

a. Involves forming a bond between the carboxyl group of the preceding amino acid to the amino group of the subsequent amino acid

b. Peptide bond forms by a condensation reaction in which a water molecule is eliminated

2. A chain of just two amino acids is a dipeptide

3. A short sequence of amino acids is considered a peptide, with longer chains being called polypeptides

4. When joined in a series of peptide bonds, the amino acids are called “residues”

5. Protein primary structure is divided into two components

a. The polypeptide backbone (has same composition for all proteins)b. Variable side chains of the amino acids

6. The arrangement of amino acids, with their distinct side chains gives each protein its characteristic structure and function

D. Protein Structure: Secondary Structure Introduction

1. Secondary structure arises due to interactions of amino acids with their neighbors (specifically the amino acid side chains)

a. Some secondary structures can be stabilized by backbone interactionsb. Interactions often stabilized by hydrogen bondingc. Can depend on disulfide bridgesd. Can depend on van der Waals forcese. Can also be stabilized by hydrophobic interactions

2. There are three types of secondary structure found in proteins

a. -helixαb. -Pleated sheetΒc. Unstructured turns

D. Protein Structure: Secondary Structure- helixα

1. The right handed -helix is the most common structural motif αfound in proteins

a. Structure was derived from theoretical models by Linus Pauling and Robert Coreyb. About 30% of all residues in globular proteins are found to be part of -helicesαc. In 1960, X-ray crystallography of myoglobin confirmed the theoretical models

2. Most amino acids can participate in an -helix although αproline cannot

a. Is cyclic and thus considered helix breakingb. Proline is actually an imino acid rather than an amino acid, because of its structure it actually puts a bend in the polypeptide chain

3. -helices are stabilized by hydrogen bonding among near αneighbor amino acids with each residue being bonded to two other residues (3 ahead and 3 behind)

4. The structure has the following characteristics

a. A pitch of 5.4 A, which is the repeat distanceb. A diameter of 2.3 A c. Contains 3.6 amino acids

5. The side chains stick out from the helical core

a. In the figure side chains are not showingb. They are essentially pointing directly into and away from the figure (can’t be seen in 2D)

E. Protein Structure: Secondary Structure-The -Pleated Sheetβ

1. The -pleated sheet also goes by the name -strandβ β

2. Was the second type of secondary structure predicted form mathematical modeling by Pauling and Corey

3. Structure involves extended amino acid chains in a protein that interact by hydrogen bonding

4. The chains are packed side by side to create a pleaded or accordian-like structure with a repeat distance of 7.0 A

5. Two segments of a polypeptide chain can form two different polypeptide structures

a. Parallel structures in which both sections are aligned with the same polarityb. Antiparallel structures in which each section has opposite polarity

F. Protein Structure: Secondary Structure-Unstructured Turns

1. “Turns” connect the -helices and -pleated sheets in proteinsα β

2. Relatively short loops that do not exhibit a defined secondary structure

3. Turns are essential to the overall folding of a protein

4. Other disordered or irregular structures in proteins are normally confined to the amino and carboxyl termini

5. Can more rarely occur to loop regions within a protein or linker region connecting one or more structural domains

G. Protein Structure: Tertiary Structure Introduction

1. Tertiary structure is the folded three-dimensional shape of a polypeptide

2. Interactions within tertiary structure are stabilized by both covalent and non-covalent interactions

a. Hydrophobic interactions (non-covalent)b. Hydrogen bonding (non-covalent)

c. Disulfide bridges (covalent interaction) which can only be broken at high temperature, at acidic pH or in the presence of reducing agents

3. Hydrophobic interactions and hydrogen bonding are the most common interactions that lead to tertiary structure

4. Hydrophobic interactions occur between amino acids with non-polar R groups and act to protect the non-polar R groups from water

a. Peptide backbone will contact waterb. R-groups will be located in the interior

5. Hydrogen bonding can occur through three types of interactions

a. Backbone/backboneb. Backbone/polar R groupc. Polar R group/Polar R group

6. There are three main categories of tertiary structure

a. Globular proteinsb. Fibrous proteinsc. Membrane proteins

H. Protein Structure: Tertiary Structure – Globular Proteins

1. Proteins that adopt a roughly spherical shape are considered globular proteins

a. Most proteins in nature adopt a spherical shapeb. Most proteins will be globular

2. Most enzymes are considered globular proteins

3. Supp. Figure: Protein Structure: Tertiary Structure – Globular Proteins

J. Protein Structure: Tertiary Structure – Fibrous Proteins

1. Fibrous proteins have a long-filamentous or rod-like structure

2. Fibrous proteins as a rule function to provide structure to the nucleus, cell, or the extra-cellular matrix by forming large polymers

3. Fibrous proteins include a a number of major designs (see fig. 5.9 B-E)

a. Collagen family proteinsb. -keratins αc. Silk Fibrion

4. The collagen family proteins have the following characteristics

a. Have a triple helical arrangement of polypeptide chains as a primary characteristicb. Major component of skin, tendons, teeth and bone

5. -keratins adopt a structure composed of “coiled-coils” of -α αhelices

a. Structural components of mammalian hooves, nails and hairb. One example is actin

6. Silk Fibrion is composed of structures produced from extended antiparallel -pleated sheetsβ

7. Supp. Figure: Protein Structure: Tertiary Structure – Fibrous Proteins

K. Protein Structure: Tertiary Structure – Membrane Proteins

1. Membrane proteins, as their name suggests are proteins that are imbedded in and span membranes

a. Cell membraneb. Nuclear membrane

2. Many of these proteins either act as receptors, or form pores

3. Membrane proteins differ from other proteins in their distribution of hydrophobic amino-acids

a. Hydrophobic amino acids group together to form membrane-spanning domains

b. Membrane spanning domains must be hydrophobic to exist within the hydrophobic environment of the lipid bilayer

L. Protein Structure: Quarternary Structure

1. A functional protein can be composed of one ore more polypeptides

2. If a protein is composed of two or more polypeptides, then it has quartenary structure

3. The stabilizing forces between the multiple polypeptides are the same as those for tertiary structure

4. Each polypeptide that contributes to the quarternary structure is termed a subunit

5. Poly-peptides can be either encoded by the same gene or by different genes

a. If encoded by the same gene, then the protein will have identical subunitsb. If encoded by different genes, then the protein will have non-identical subunits

6. An example of a protein that shows quartenary structure is hemoglobin

7. Hemoglobin is composed of four peptides

a. 2 -globin peptidesαb. 2 -globin peptidesβ

8. The -globin peptides are encoded by the -globin geneα α

9. The -globin peptides are encoded by the -globin geneβ β

10. The presence of multiple peptides coming together to form the hemoglobin protein gives it quartenary structure

V. Gene Expression: Relating Genotype and Phenotype

A. Gene Expression: From Genotype to Phenotype

1. Now, let’s look at the disease sickle cell anemia

2. Sickle cell anemia is a disease in which the patient makes defective hemoglobin, as compared to an unaffected person which makes functional hemoglobin

3. Normal hemoglobin is actually a flexible protein, and allows the cell to take a bi-concave shape

4. The defective version of hemoglobin is much more rigid, such that cells take on a sickle shape

5. Sickle cell anemia shows an autosomal recessive mechanism of inheritance

6. However, the wild type allele is incompletely dominant

7. What does this mean???

8. This means that the individual with two wild-type alleles is completely healthy

9. This means that the individual that is a heterozygote will not have any disease symptoms, although he/she is more resistant to malaria

10. Patients with sickle cell anemia have symptoms of anemia

a. Shortness of breathb. Dizzinessc. Headached. Coldness of hands and feete. Pale skin

11. Patients with sickle cell anemia have symptoms of pain

a. Pain is called sickle cell crisisb. Pain often affects the bones lungs abdomen and joints

12. On a physiological level, those with -globin mutation have βred blood cells with a sickle shape, which can clump and block blood flow through small blood vessels

13. Unaffected individuals have red blood cells with a bi-concave shape

14. The -globin gene is on chromosome 11β

15. There are two main alleles for the -globin geneβ

a. Wild-type allele b. Mutant allele

16. An allele is a term in molecular biology that refers to a two or more alternative sequence variations of a specific gene

a. Wild-type variation is the most common variation in the populationb. All other variations are considered mutant (have at least one change in sequence)

17. The wild-type allele is designated by the letter A

a. The wild-type allele is dominantb. The wild type allele has the following sequence in the amino terminal region: 5’ - GTG CAC CTG ACT CCT GAG GAG - 3’

18. The mutant allele is designated by the letter s

a. This allele is recessive to the wild-typeb. The mutant allele has the following sequence in the amino terminal region: 5’ - GTG CAC CTG ACT CCT GTG GAG - 3’

19. It should be noted that base 17 is switched from an adenine to a thymine

20. This switch results in an amino acid change from a glutamate (acidic residue) to a valine (non-polar residue)

B. Gene Expression: From Genotype to Phenotype (Homozygotes)

1. To relate this to an actual medical situation, each individual will have two copies of the -globin geneβ

2. For most (if not all) of us have no symptoms of sickle cell anemia and so therefore, we have the genotype AA

3. This means that we have only the A allele. This A allele has the normal sequence GTG CAC CTG ACT CCT GAG GAG, and will produce -globin protein with the glutamic acid onlyβ

4. We all have normal hemoglobin, and thus will have erythrocytes (RBCs) with a normal bi-concave shape

C. Gene Expression: From Genotype to Phenotype-The Homozygous Sickle Cell Individual

1. Individuals with full blown sickle cell anemia have the genotype ss

2. These individuals have two copies of the -globin gene-each βwith the (s) mutant allele GTG CAC CTG ACT CCT GTG GAG

3. Therefore, from both copies they will make the defective -βglobin protein with the valine instead of the glutamic acid

4. In the end, they will end up with sickle red blood cells, and will have disease symptoms

D. Gene Expression: From Genotype to Phenotype-The Heterozygotes

1. Heterozygotes (genotype As), when it comes to sickle cell anemia are an interesting group

2. The wild-type allele is incompletely dominant – with these alleles for -globin the effects of both are seenβ

3. They have one copy of the -globin gene that has the normal βsequence GTG CAC CTG ACT CCT GAG GAG

4. They also have one copy of the -globin gene that has the βmutant sequence GTG CAC CTG ACT CCT GTG GAG

5. From one of the copies they will produce the normal -globin βprotein, with the glutamic acid, and could make potentially normal hemoglobin

6. From the other copy, they will produce the mutant (disease) causing -globin protein with the valine, and could make sickle βcell hemoglobin

7. In this case only 50% of the protein produced is normal, and 50% is mutant

8. Since As heterozygotes do not show a sickle cell phenotype, then as long as you make 50% normal -globin protein, you get βenough normal hemoglobin to be healthy

E. Gene Expression: From Genotype To Phenotypes-Genes That Show Normal Dominance

1. The CFTR gene is implicated in the development of the disease cystic fibrosis

2. CFTR stands for cystic fibrosis transmembrane conductance regulator

3. The CFTR gene encodes the CFTR protein which functions as a chloride channel (Pore)

4. Given the fact that the CFTR protein is a channel (pore), it is a transmembrane protein

5. Patients who have cystic fibrosis have very debilitating symptoms and a reduced life span

6. The symptoms include

a. Thick viscous mucus in lungsb. Repeated infectionsc. Chronic pneumoniasd. Chronic cough with blood streakinge. Wheezingf. Excessively salty sweatg. Fatigueh. Reduced male fertility

7. Drugs today have allowed patients to live longer more productive lives

8. The CFTR gene is found on the long arm of chromosome 7

9. Most patients with CF have a mutation that deletes codon 508 from the CFTR gene

10. This codon removes a TTT codon (which in the corresponding mRNA is UUU)

11. Therefore a Phenylalanine is deleted from the protein-rendering the function of the chloride channel defective

12. The mutant CFTR allele in which codon 508 is deleted is recessive, and the normal, wild-type allele is completely dominant

13. Therefore, a patient needs to have two copies of the mutant CFTR allele to have the disorder

14. Now we know in the case where an individual is homozygous for the wild-type allele, they should only produce the normal CFTR protein

15. We know in the case of the patient with the CF disease that they have two copies of the mutant allele and will only produce defective CFTR protein

16. What is the molecular explanation for why the heterozygotes have a normal phenotype

17. A heterozygote will have one copy of the normal allele, from which normal protein will be produced

18. A heterozygote will have one copy of the mutant allele, which will produce the defective protein

19. Therefore, this persons cells should produce 50% normal and 50% defective protein

20. As long as there is 50% normal, functional protein, the individual is healthy