Chapter 2Nucleic Acids Convey Information: A View Of Gene Expression

I. IntroductionA. Introduction: Molecules Are Necessary For Life

“Life…..is a relationship between molecules”–Linus Pauling

1. The idea that molecules could carry information began in the 1800s with the work of Friedrich Meischner

2. Work continued showing that DNA is the prime genetic molecule

a. Avery, McCleod and McCartyb. Hershey and Chase

3. 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

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

5. “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

B. Introduction: Understanding How Molecules Affect Human Health

1. Understanding molecular basis of gene expression is important to modern medicine as it allows for treatment of single gene disorders

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

b. Cystic fibrosisc. Muscular dystrophyd. Tay-Sachs Disease e. Sickle Cell Anemia

2. Understanding the molecular basis of gene expression allows for the development of treatment of complex genetic disorders

a. Cancerb. Alzheimer’s diseasec. Diabetes

3. Understanding gene expression allows for the development of vaccines as well as treatments to pathogens

C. Introduction: Understanding Gene Expression

1. Two key questions arose when it came to the study of DNA as the prime genetic molecule

a. How genes are expressed (How does information in DNA encode proteins)b. How is the DNA replicated

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

a. Enzymes (many of which are multi-subunit enzymes)b. Proteins that become part of larger structures (ex. Ribosome)

II. The Central Dogma of Molecular Biology

A. The Central Dogma: Introduction

1. The flow of information in gene expression from the prime genetic molecule

a. Is termed as the “Central Dogma of molecular Biology”b. Coined by Francis Crickc .“Once information is passed into protein it cannot get out again” –Francis Crick

2. The central dogma of Molecular Biology: DNA mRNA Protein

3. A gene can encode for one of two types of products

a. Protein (in most cases)b. Non-coding RNA

4. For protein coding genes

a. Protein is the terminal productb. Carries out the function of the gene

5. For genes encoding RNA (RNA carries out function of gene)

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 transcription

a. DNA is used as a template to produce RNAb. If the RNA produced is a messenger RNA, then this is not the terminal step in gene expressionc. For all other RNAs produced, this is the terminal step

3. The arrow that flows from RNA to protein indicates translation

a. Only 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

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. Genes have a specific structure and have different regions

2. DNA is a double-stranded molecule

a. Top strand is considered the Watson Strandb. Bottom strand is considered the Crick Strand

3. Genes can go in both directions along a DNA molecule

a. Left to right (5’ 3’)b. Right to (5 ‘ 3’)

4. Each gene has both a coding and non-coding strand and directionality of the gene is determined by the direction the coding strand goes

5. The coding strand typically has the same sequence as the mRNA that will be produced

a. Contains thymine instead of uracilb. For genes located on the Watson strand, the coding strand goes 5’ 3’ left to rightc. For genes located on the Crick strand, the coding strand is the one that goes 5’ 3’ right to left

6. The non-coding strand is complementary to the mRNA that will be produced and serves as a template for mRNA production

a. For genes located on the Watson strand, the non-coding strand goes 5’ 3’ right to leftb. For genes located on the Crick strand, the non-coding strand goes 5’3’ left to right

7. 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

8. Genes are numbered based on where transcription starts

a. The base pair where transcription starts is designated number 1 (based off of the coding strand)

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

9. The promoter is generally close to the actual gene itself and is found upstream to the transcriptional unit

a. Upstream boundary can extend up to -100 bpb. Downstream can extend up to +20 bp-+30 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

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. Supplemental Figure: Basic Mechanisms of Gene Expression: Transcription

C. 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 ribosome translates the mRNA sequence

a. Three bases at a time into a single amino acid b. A protein is made up of a polymer of amino acids

E. 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 strandb. 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 proteins

b. 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

F. 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 in the 5’ 3’ direction 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. Multiple nitrogenous bases must encode for a single amino acid

a. Only 16 possible two base codes and 20 total amino acidsb. More than two bases must encode a single amino acid

7. There are 64 possible three base codes and a total of 20 amino acids

a. Three base codes are necessary for encoding a single amino acid b. Each 3 base code is called a codonc. Each amino acid can be encoded by multiple codons

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

9. Take an mRNA that has the sequence 5-P-CCC UAG GCA CUC AUG CCC UUU GCU GGG UAA CUG UCG CUA AAC AAU AAA-OH 3’ and translate it

10. The start codon for translation is AUG

11. 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

12. 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. Placement of the start codon establishes the remaining codons by moving 3 bases each time towards the 3’end of the mRNA

13. 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

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

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

15. The ribosome will translate the mRNA in the 5’ 3’

16. Synthesis of the new peptide occurs from the amino terminus-to the carboxy terminus

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

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

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

19. Therefore, we have a protein with the sequence of NH2-MET-PRO-PHE-ALA-GLY-COOH

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

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

3. This observation 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.

H. 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

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

1. One outstanding question was how each codon coded for a specific amino acid

2. Idea #1: 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 (dependent on the R-Group)

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

a. Most amino acids have non-polar side chainsb. Several amino acids have very similar R-groups

4. Idea 2: 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

5. 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 translation

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

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

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

IV. Protein Structure

A. Protein Function: Introduction1. Proteins have a variety of diverse functions in cells

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

2. The process of gene expression is not complete after translation

a. Protein must fold properlyb. Proper folding is important for function

3. Protein folding occurs co-translationally

4. The proper folded structure is dependent on two factors

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

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

a. Chaperones required for folding of most proteinsb. Chaperones work co-translationally

6. There are four levels of proteins structure

a. Primaryb. Secondary c. Tertiaryd. Quartenary

7. Amino acid chemistry has a great influence on protein folding

8. 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

9. 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

10. 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

B. Protein Structure: Primary Structure

1. The primary sequence is really just the sequence of amino acids present in a protein

2. Primary structure is formed by joining amino acid together via peptide bonds

a. Form a bond between the carboxyl group of the preceding amino acid to the amino group of the subsequent amino acidb. Peptide bond forms by a condensation reaction

3. Nomenclature for a growing protein

a. A chain of just two amino acids is a dipeptideb. Short chains of amino acids is considered a peptide, c. Longer chains being called polypeptidesd. When joined in a series of peptide bonds, the amino acids are called “residues”

4. Proteins, just like DNA and RNA strands have a specific polarity

a. Amino acid structure is not symmetricalb. The terminus with the free amino group is considered the amino terminusc. The terminus with the free carboxyl group is considered the carboxyl terminus

5. Note: Each of the 20 amino acids has a different R group

a. Non-polarb. Polarc. Charged (acidic and basic)

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

7. Each type of R-group allows the amino acid to participate in a variety of different interactions to allow for more complex folding

C. Protein Structure: Secondary Structure Introduction

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

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

a. -helixαb. -Pleated sheetΒc. Unstructured turns

3. Secondary structures are stabilized by a combination of the following types of interactions involving the peptide backbone

a. Interactions often stabilized by hydrogen bondingb. Can depend on disulfide bridgesc. Can depend on van der Waals forcesd. Can also be stabilized by hydrophobic interactions

D. Protein Structure: Secondary Structure- helixα

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

a. Structure derived from theoretical models by Linus Pauling and Robert Corey

b. In 1960, X-ray crystallography of myoglobin confirmed the theoretical models

2. All amino acids can participate in an -helix except proline α

a. Cyclic and thus considered helix breakingb. Actually an imino acid rather than an amino acid, and will put 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

8. Amino acid 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. Unstructured “Turns” connect the -helices and -pleated α βsheets in proteins

a. Relatively short loops that do not exhibit a defined secondary structureb. Turns are essential to the overall folding of a protein

2. 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. The folded three-dimensional shape of a polypeptide is tertiary structure

2. Interactions within tertiary structure are stabilized by the following interactions

a. Hydrophobic interactions (non-covalent) (more common)b. Hydrogen bonding (non-covalent) (more common)c. Disulfide bridges (covalent interaction) which can only be broken at high temperature, at acidic pH or in the presence of reducing agents (less common)

3. Hydrophobic interactions occur between amino acids with non-polar R groups

a. Protect the non-polar R groups from waterb. Peptide backbone will contact waterc. R-groups will be located in the interior

4. Hydrogen bonding can occur through three types of interactions

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

5. 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. Supplemental 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 by forming large polymers

a. Nucleusb. Cellc. Extra-cellular matrix

3. Fibrous proteins include a a number of major designs

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

K. Protein Structure: Tertiary Structure – Membrane Proteins

1. Membrane proteins are imbedded in and span membranes

a. Cell membraneb. Nuclear membrane

2. Membrane proteins can have the following functions

a. Act as receptorsb. 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 domainsb. Membrane spanning domains must be hydrophobic to exist within the hydrophobic environment of the lipid bilayer

L. Protein Structure: Quarternary Structure

1. A protein is composed of two or more polypeptides has quartenary structure

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

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

4. 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

5. Hemoglobin has quaternary structure is composed of four peptides

a. 2 -globin peptidesαb. 2 -globin peptidesβ

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

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

V. Molecular Pathology

A. Gene Expression: From Genotype to Phenotype

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

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

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

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

5. However, the wild type allele is incompletely dominant

6. An individual with two wild-type alleles is completely healthy

7. A heterozygote will have some mild disease symptoms, and he/she is more resistant to malaria

8. Patients with sickle cell anemia have symptoms of anemia

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

9. 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

10. 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

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

12. The -globin gene is on chromosome 11β

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

a. Wild-type allele (A)b. Mutant allele (s)

14. Allele: a term in molecular biology that refers to a sequence variation 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)

15. The wild-type allele:

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’

16. The mutant allele:

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’

17. Base 17 is switched from an adenine to a thymine

18. 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 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)

a. They have one copy of the -globin gene that has the βnormal sequence GTG CAC CTG ACT CCT GAG GAGb. They also have one copy of the -globin gene that has βthe mutant sequence GTG CAC CTG ACT CCT GTG GAG

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

a. From one of the copies they will produce the normal -βglobin protein, with the glutamic acid, and could make potentially normal hemoglobinb. From the other copy, they will produce the mutant (disease) causing -globin protein with the valine, and βcould make sickle cell hemoglobin

3. Only 50% of the protein produced is normal, and 50% is mutant and they show an intermediate phenotype

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. Fatiguee. 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