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Pre-7.01 Getting up to Speed in Biology

Last updated Feb 19, 2023

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About this Course

This self-paced course was originally designed to help prepare incoming MIT students for their first Introductory Biology course (known at MIT as 7.01). It will also be useful for anyone preparing to take an equivalent college-level introductory biology class elsewhere.


# Lecture 1 - Molecules of Life

# Lecture 1.1 - Representing Molecules

# Representing Molecules

# Chemical Formula

Each C, H, and O atom is counted. For example…

C 4 H 9 O H

# Full Structure

All atoms and bonds are shown. For example…

Butane is represnted as a linear chain of four C’s that are connected by dashes. Each dash represents a covalent bond. Each C has an H above it and below it, each of which is connected to the appropriate C by a dash. The C at the leftmost end of the chain has an additional H connected by a dash to its left. The C at the rightmost end of the chain is connected by a dash to an O that is in turn connected by a dash to an H.

# Line-angle Structure

Below is butane. Carbons are at the corners or ends, and hydrogens are not shown.

Butane is represented as a zig-zag line with 3 obtuse angles with an O H group at the right end of the line. smiles forward

The locations of the carbons are shown below with blue dots.

Again, butane is represented as a zig-zag line with 3 obtuse angles with an O H group at the right end of the line. Additionally, there are blue dots at the leftmost end of the line and at all three verticies at the angles.

Now, the locations of the hydrogens are shown below with red dots.
Because C has a valency of FOUR (meaning that each C makes four bonds), you can figure out location of H atoms.

Again, butane is represented as a zig-zag line with 3 obtuse angles with an O H group at the right end of the line. Additionally, there are blue dots at the leftmost end of the line and at all three verticies at the angles. There are two red dots at the periphery of of the blue dots at a vertex. There are three red dots at the periphery of the blue dot at the leftmost end of the line.

# Lecture 1.2 - Polar and Non-polar Molecules

# Dipoles

An unequal electron distribution is called a dipole.
Dipoles are often shown by a T-arrow or a δ. The dipoles are shown on the figure in red. The overall dipoles (which are a vector sum of the individual dipoles) are shown in blue.

# For CO2…

The C=O bond is POLAR since O is more electronegative than C (attracts electrons more), BUT the vector sum of the dipoles is equal to zero, so CO2 is NON-POLAR overall.

# For H2O…

The O-H bond is POLAR since O is more electronegative than H. The vector sum of dipoles is NOT equal to zero, so H2O is POLAR overall.

# Example: Propane v. Propanol

This image is of propane: CCC

This is propane.
Chemical formula: C3H8
Bonds: NON-POLAR, since C and H share electrons equally
Molecule: NON-POLAR

This image is of propanol: CCCO. The O is labelled with a delta minus to indicate that it has a partial negative charge symbol. The C and H that are attached to this O atom are each labelled with a delta plus symbol to indicate that they each have partial positive charges.

This is propanol.
Chemical formula: C3H7OH or C3H8O
Bonds: C-O and O-H bonds are POLAR, since O is more electronegative (attracts electrons more) than C or H
Molecule: POLAR, the dipole vectors do not cancel out.

# Lecture 1.3 - Types of Bonds

# Ionic Bond

An ionic bond is an electrostatic interaction between oppositely charged ions (atoms that have lost/gained 1 or more electrons).
Strong interaction

This image is of acetate CC(=O)[O-] and of a sodium ion [Na+]. The sodium ion is positioned next to the negatively-charged O atom of the acetate. The negatively charged oxygen and the positively charged sodium are circled to indicate that they interact via an ionic bond.

# Covalent Bond

A covalent bond consists of one or more pairs of electrons are shared by two atoms.
Strong interaction

This image is of acetate CC(=O)[O-] and of a sodium ion [Na+]. The sodium ion is positioned next to the negatively-charged O atom of the acetate. Double bond between the central carbon and the neutral charge oxygen is circled to indicate that the interaction is a type of covalent bond.

# Hydrogen Bond

A hydrogen bond forms between polar molecules where an H has a partial positive charge (induced by an electronegative atom) and is therefore attracted to a different electronegative atom.
Medium-strong interaction

This image is of adenine C1=NC2=NC=NC(=C2N1)N and thymine CC1=CNC(=O)NC1=O. One of the hydrogens from adenine’s amine group is aligned with an oxygens of one thymine’s adehyde groups. There is a dashed line between the H and the O. This dashed line is circled to indicate that the interaction is a hydrogen bond. That H atom is also labelled with a delta plus symbol. The N of that amine group and the O atom are both labelled with delta minus symbols. Additionally, there is a dashed line between the H atom of thymine’s amine group and one of the N atoms of the six-membered adenine ring. This dashed line is also circled to indicate that the interaction is a hydrogen bond. This H atom is also labelled with a delta plus symbol. The N atom that the H atom forms a H bond with and the N atom from the amine group are labelled delta minus symbols.

# Non-polar/van Der Waals Bond

This type of bond is between nonpolar molecules, with transient unequal charge distribution (dipoles).
Weak interaction

This image is of two molecules of pentane CCCCC. There is a dashed oval encompassing part of each molecule. The oval is labelled with a delta plus symbol at one end and a delta minus symbol at the other end. The oval indicates that transient van der Waals interactions are forming between the atoms of the two molecules.

# Lecture 1.4 - Recognizing Macromolecules

# Lipids

Lipids are non-polar and hydrophobic.

This image is of trans-oleic acid: CCCCCCCCC=CCCCCCCCC(=O)O

# Carbohydrates

Carbohydrates follow the formula CH2O and are polar.

This image is of glucose: C(C(C(C(C(C=O)O)O)O)O)O

# Proteins

Proteins are made up of amino acid monomers, which consist of an α-carbon (circled in red) bound to an NH2 group, a COOH group, and a side chain, or R group, (circled in blue). The R group can be polar or non-polar

This image is of valine: CC(C)C(C(=O)O)N

# Nucleic Acids

Nucleic acids are built from nucleotide monomers, which consist of a phosphate group (circled in green), a sugar (circled in orange), and a base (circled in purple). Nucleotides are polar. ^86dbdb

This image is of guanosine monophosphate: C1=NC2=C(N1C3C(C(C(O3)COP(=O)(O)O)O)O)N=C(NC2=O)N

# Lecture 1.5 - Nucleic Acid Polarity

# Anatomy of a Nucleic Acid

Above is a diagram of a nucleic acid polymer.
Recall that each nucleotide is composed of a sugar, phosphate and base.
S = Sugar
P = Phosphate
B= Base.

The strand is polymerized from the 5’ end to the 3’ end. The nucleotide at the 5’ end has a free phosphate, and the nucleotide at the 3’ end has a free 3’OH group on its sugar.

# Nomenclature

S-P backbone is not written, just the bases and the polarity of the strand.
ALWAYS write 5’ and 3’ on each nucleic acid strand!

In other words, written nucleic acids take the following form:
5’B1B2B3B4B5B63’

For example:
5’GAATCC3’

# Polarity of Synthesis

Nucleic acids are made from 5’P to 3’OH. This means that the –3’OH end is the growing end, i.e. receives the incoming nucleotide, as shown below.

Base order = INFORMATION
Polarity = 5’ and 3’ ends show first to last nucleotide added and confer a direction to read information

# Lecture 1.6 - Protein Polarity

# Anatomy of an Amino Acid Polymer

The diagram above shows the synthesis of a dipeptide from two amino acid monomers.
The alpha-carbons have been circled in orange.
The peptide bond that forms upon synthesis is circled in blue.

The polypeptide is polymerized from the N end to the C end. The amino acid at the N end has a free NH2 group and the amino acid at the C end has a free COOH group.

# Nomenclature

Proteins are written with three or one letter amino acid code from N to C.
ALWAYS write N and C on each protein!

For example:

OR, from the diagram above:

# Polarity of Synthesis

A polypeptide chain is synthesized from N to C. Therefore, the –C end is the growing end (for addition of next amino acid).
Amino acid order = INFORMATION
Polarity = N and C ends show first to last amino acid added

# Lecture 2 - The Cell and how it Works

# Lecture 2.1 - Condensation and Hydrolysis

# Anabolism v. Catabolism

Anabolism is a form of metabolism that consumes energy and by which covalent bonds are formed.
Catabolism is a form of metabolism that produces energy and by which covalent bonds are broken.

# Condensation Reactions

A condensation reaction is one that forms covalent bonds and produces water as a product. These reactions are anabolic.

A-OH + H-B → A-B + H2O
(This reaction is also included in the figure at end.)

# Hydrolysis Reactions

Hydrolysis reactions break covalent bonds by consuming a water molecule and dividing its atoms between separate molecules. These reactions are catabolic.

A-B + H2O → A-OH + H-B
(This reaction is also included in the figure below.)

# Lecture 2.2 - Free Energy and Reaction Kinetics

# $$

\Delta G = \Delta H - T \Delta S
$$

where:

1
2
3
4
  G = free (usable) energy  
  H = enthalpy (total energy)  
  T = temperature  
  S = entropy (unusable energy)

# Free Energy (ΔG)

ΔG = ΔH - TΔS
H = Enthalpy
S = Entropy
T= Temperature

# Exergonic Reactions

Exergonic reactions release energy, so that ΔG < 0 and the reaction is spontaneous.

This image depicts a conceptual graph. The graph is titled Exergonic Reaction. The y axis label is Free Energy. The x axis label is Course of Reaction. There is one curve plotted on the graph that extends from a plateau at the top left of the plot and then curves down to a plateau at the bottom right of the plot. The value of the curve at the top left is labeled Reactants. The value of the curve at the bottom right is labeled Products. Finally, the value of the Energy Released, or delta G, is labeled as the difference between the value labeled Reactants and the value labeled Products.

# Endergonic Reactions

Endergonic reactions require energy to proceed, so that ΔG > 0 and the reaction is non-spontaneous.

This image depicts a conceptual graph. The graph is titled Endergonic Reaction. The y axis label is Free Energy. The x axis label is Course of Reaction. There is one curve plotted on the graph that extends from a plateau at the bottom left of the plot and then curves up to a plateau at the top right of the plot. The value of the curve at the bottom left is labeled Reactants. The value of the curve at the top right is labeled Products. Finally, the value of the Energy Released, or delta G, is labeled as the difference between the value labeled Reactants and the value labeled Products.

# Equilibrium

If ΔG = 0, the reaction is at equilibrium such that the rate fo the forward reaction = the rate of the backwards reaction.

# Lecture 2.3 - Pathways

# Biochemical Pathways

Biochemical pathways include activation (positive control), represented by a pointed arrow (→),
and inhibition (negative control), represented by a T-bar arrow.
Feedback serves to regulate product levels within a pathway.

# Lecture 2.4 - Organelles

# What is a Cell?

A cell is the smallest unit of life that is bound by a membrane and can self-replicate.

There are two types of cells: eukaryotic cells (such as the cells that make up plants and animals) and prokaryotic cells (such as bacteria cells).
An organism that is made up of one or more eukaryotic cells is called a eukaryote, and an organism whose cells are prokaryotic is called a prokaryote. In this way, plants and animals can be referred to as eukaryotes and bacteria are examples of prokaryotes.

# Organelles

An organelle is defined as a specialized structure within a cell that serves a specific function.

The chart below lists eukaryotic organelles with a short summary of their functions. Pay particular note to the nucleus, mitochondrion (plural: mitochondria), and ribosome.

Common Organelles of Eukaryotic Cells
NameFunction
Organelles with membranes
NucleusProtecting, controlling access to DNA
Endoplasmic Reticulum (ER)Routing, modifying new polypeptide chains; synthesizing lipids; other tasks
Golgi bodyModifying polypeptide chains; sorting, shipping proteins and lipids
VesicleTransporting, storing, or digesting substances in the cell, other functions
MitochondrionMaking ATP by breaking down sugars
ChloroplastMaking sugars in plants and some protists
LysosomeIntracellular digestion
PeroxisomeInactivation of toxins
VacuoleStorage
Organelles without membranes
RibosomeAssembling polypeptide chains
CentrioleAnchor for the cytoskeleton

# Lecture 2.5 - Cell Division

# Mitosis

Mitosis is the process of cell division by which 2 daughter cells are produced from 1 mother cell.
The daughter cells have the same ploidy and are genetically identical to both each other and the mother cell.

This diagram depicts the process of mitosis in which a cell with a pair of homologous chromosomes replicates. First, there is a cell, represented as a circle, that has two chromosomes. These chromosomes are each represented as a line in the cell. One chromosome is labelled A 1 and the other is labeled A 2. Then, an arrow labelled D N A replication points from this first cell diagram to a second cell diagram. This second circle contains four lines that represent the two copies of the two original chromosomes. The two lines labelled A 1 are paired and the other two lines labelled A 2 are paired. There are two additional cells that are drawn with an addition sign in between them. There are two arrows that stem from the second cell diagram, and each one points to one of these addtitional cells to indicate that the cell has divided into two new cells. The two new cells are identical with each other and the first diagram of the original cell in that they contain one A 1 chromosome and one A 2 chromosome. That is to say that each daughter cell is diploid and identical to the mother cell.

# Meiosis

Meiosis is the process of cell division by which 4 gametes are produced from 1 mother cell.
The gametes’ ploidies are half of the mother cell’s ploidy. The gametes are NOT genetically identical to each other NOR to the mother cell.

Gametes are cells with half of the normal ploidy (haploid for humans). Examples of gametes are eggs (ova) and sperm cells.

This diagram depicts a cell with a pair of homologous chromosomes undergoing meiosis. First, there is a cell, represented as a circle, that has two chromosomes. These chromosomes are each represented as a line in the cell. One chromosome is labelled A 1 and the other is labeled A 2. Then, an arrow labelled D N A replication points from this first cell diagram to a second cell diagram. This second circle contains four lines that represent the two copies of the two original chromosomes. The two lines labelled A 1 are paired and the other two lines labelled A 2 are paired. There are two arrows that stem from the second cell diagram, and each one points to a new cell to indicate that the original cell has divide into two new cells. In this step, known as Meiosis I, the homologous chromosomes separate. Therefore, one of these new cells contains the two paired A 1 chromosomes and the other new cell contains the two paired A 2 chromosomes. The next step is labelled Meiosis II and is when sister chromatids separate. Each of the two cells produced from Meiosis I divides to produce two new cells. Each of these new cells contains one copy of one chromosome. Therefore, a total of four cells are produced. Two of these product cells have a single copy of the A 1 chromosome, and two of these product cells have a single copy of the A 2 chromosome. Thus, the products of meiosis are 4 haploid gametes.

# Lecture 3 - Information Transfer in Biology

# Lecture 3.1 - DNA Rules

# DNA Polarity

Recall that DNA strands have a 3’ end and a 5’ end and that new nucleotides are added to the free -OH group of the 3’ end.

# Base Pairing

For DNA, there are 4 nucleotide bases: Guanine (G), Cytosine (C), Adenine (A), and Thymine (T).
G pairs with C between complimentary strands via the formation of 3 hydrogen bonds.
Similarly, A pairs with T via the formation of 2 hydrogen bonds.

For RNA, the bases are the same, except Uracil (U) is used instead of Thymine.
In RNA, A pairs with U, and they form 2 hydrogen bonds.
(Therefore, the RNA bases are: G, C, A, U.)

# Lecture 3.2 - DNA Replication

# The Steps of DNA Replication

Step 1. The two strands of the original DNA separate.

This image depicts double stranded D N A being separated into two pieces of single stranded D N A. The top strand represented as a black arrow that starts at the 5 prime end and ends at the 3 prime end. A sequence is given in the middle of the arrow. The sequence is G A T T A C A. The bottom strand is complementary and antiparallel. The bottom strand also represented as a black arrow that goes from the 5 prime end to the 3 prime end; the 3 prime end is on the left and the 5 prime end is on the right. The sequence of this strand is C T A A T G T. This double stranded D N A has two thick arrows that show it being separated into two single stranded molecules. The top strand is shown on the left and the bottom strand is shown on the right.

Step 2. RNA primers attach to the separated DNA strands.

This image depicts R N A primers attaching to the single stranded D N A strands. At the left, the D N A sequence is represented as a black arrow that starts at the 5 prime end and ends at the 3 prime end. This strand has the following sequence: G A T T A C A. The D N A strand is labelled as a template. There is a short gray arrow that is aligned at the bottom right of the D N A template. This gray arrow is labelled as an R N A primer. The gray arrow goes from the 5 prime end to the 3 prime end but the 3 prime end is on the left and the 5 prime end is on the right. In this way, the 5 prime end of the R N A primer is aligned with and attached to the 3 prime end of the D N A template. At the right, the other D N A sequence is represented as a black arrow that goes from the 5 prime end to the 3 prime end; the 3 prime end is on the left and the 5 prime end is on the right. This strand has the following sequence: C T A A T G T. The D N A strand is labelled as a template. Just as before, there is a short gray arrow that is aligned at the top left of the D N A template. This gray arrow is also labelled as an R N A primer. The gray arrow goes from the 5 prime end at the left to the 3 prime end at the right. In this way, the 5 prime end of the R N A primer is aligned with and attached to the 3 prime end of the D N A template.

Step 3. The DNA is replicated off of the 3’ end of the RNA primers.

This image depicts the synthesis of new D N A via extention of the R N A primers that are attached to the single stranded D N A strands. The diagram starts with a replication of the figure from step 2. The description of this first part of the diagram is repeated here. At the left, the D N A sequence is represented as a black arrow that starts at the 5 prime end and ends at the 3 prime end. This strand has the following sequence: G A T T A C A. The D N A strand is labelled as a template. There is a short gray arrow that is aligned at the bottom right of the D N A template. This gray arrow is labelled as an R N A primer. The gray arrow goes from the 5 prime end to the 3 prime end but the 3 prime end is on the left and the 5 prime end is on the right. In this way, the 5 prime end of the R N A primer is aligned with and attached to the 3 prime end of the D N A template. At the right, the other D N A sequence is represented as a black arrow that goes from the 5 prime end to the 3 prime end; the 3 prime end is on the left and the 5 prime end is on the right. This strand has the following sequence: C T A A T G T. The D N A strand is labelled as a template. Just as before, there is a short gray arrow that is aligned at the top left of the D N A template. This gray arrow is also labelled as an R N A primer. The gray arrow goes from the 5 prime end at the left to the 3 prime end at the right. In this way, the 5 prime end of the R N A primer is aligned with and attached to the 3 prime end of the D N A template. Now, there are two thick arrows that show each of these D N A and R N A pairs being converted to new double stranded D N A molecules. The product from the left template is a double stranded D N A where the top strand is black and bears the sequence 5 prime G A T T A C A 3 prime. This strand was from the original double stranded D N A. The bottom strand is gray and bears the sequence 3 prime C T A A T G T 5 prime. This strand is newly synthesized. The product from the right template is a double stranded D N A where the top strand is gray and bears the sequence 5 prime G A T T A C A 3 prime. This strand is newly synthesized. The bottom strand is black and bears the sequence 3 prime C T A A T G T 5 prime. This strand was from the original double stranded D N A.

Thus, the result of DNA replication is two semiconservative, identical DNA molecules.

# Lecture 3.3 - Transcription

# Transcription

An RNA molecule is synthesized from one strand of a dsDNA; this strand used for synthesis is called the template strand.

Like DNA, RNA is synthesized 5’ to 3'.

Once finished, the RNA transcript dissociates from the template strand, and the two DNA strands hybridize again to form dsDNA.

This figure shows the process of transcription. The figure is divided by dashed lines into three sections. The first section is captioned: gene (d s D N A). Here, a double stranded D N A is shown as a pair of horizontal black lines. The top line is labelled as the non-template strand with a 5 prime label on its left end and a 3 prime label on its right end. The bottom line is labelled as the template strand with a 3 prime label on its left end and a 5 prime label on its right end. There is a thick arrow spanning a dashed line to mark the transition to the second section. The two DNA strands are repeated in the second section, although now they are separated by a larger distance. There is a gray arrow that points from a 5 prime label on its left end to a 3 prime label on its right end. This gray arrow is aligned on just above the bottom D N A strand, which is the template strand and runs from 3 prime on the left to 5 prime on the right. Below the bottom D N A strand is a bracket that spans the length and location of the gray arrow. The bracket is labelled as the transcribed region. Furthermore, the left side of the bracket is labelled as the start site and the right side of the bracket is labelled as the stop site. This second section has the following caption: The template is copied from 3 prime to 5 prime, and the RNA is synthesized 5 prime to 3 prime. There is another thick arrow spanning a dashed line to mark the transition to the third and final section. This section is captioned: R N A (transcript). The gray arrow from before is copied in this third section as a gray line and is labelled 5 prime on its left end and 3 prime on its right end.

# Lecture 3.4 - Translation

# Translation

Translation is the process by which RNA is decoded by the ribosome to produce a polypeptide.

codon is a set of 3 consecutive nucleotides that together code for an amino acid.

Adaptor RNA molecules called tRNAs bring amino acids to the ribosome.
The tRNAs have anticodons that pair (antiparallelly) with the codon to ensure that the correct amino acid is being added.

# The Codon Chart

The codon chart is used to determine which amino acid corresponds to which codon. Notice that some amino acids have multiple codons but each codon only codes for one amino acid.

Here is a codon chart.

Here is a codon chart that is used to identify the amino acids that correspond to each R N A codon. The chart has 4 rows and 4 columns, both of which correspond to the 4 possible nucleotides of R N A (which are U, C, A, G). The rows correspond to the first letter of the codon (at the 5 prime end of the sequence) and is selected first. Then, a column is chosen; the columns correspond to the second letter of the codon. Then, the 4 possible codons (corresponding to the 4 possible options for the third letter (at the 3 prime end of the sequence)) are listed in the selected box with the corrresponding amino acid written next to the codon. Each amino acid is notated with its three-letter abreviation. The codons of all the boxes will now be enlisted in the order (from left to right) written on the chart: U U U = Phenylalanine or P h e, U U C = Phenylalanine or P h e, U U A = Leucine or L e u, U U G = Leucine or L e u, U C U = Serine or S e r, U C C = Serine or S e r, U C A = Serine or S e r, U C G = Serine or S e r, U A U = Tyrosine or T y r, U A C = Tyrosine or T y r, U A A = STOP, U A G = STOP, U G U = Cysteine or C y s, U G C = Cysteine or C y s, U G A = STOP, U G G = Tryptophan or T r p, C U U = Leucine or L e u, C U C = Leucine or L e u, C U A = Leucine or L e u, C U G = Leucine or L e u, C U A = Leucine or L e u, C U G = Leucine or L e u, C C U = Proline or P r o, C C C = Proline or P r o, C C A = Pro, C C G = Proline or P r o, C A U = Histidine or H i s, C A C = Histidine or H i s, C A A = Glutamine or G l n, C A G = Glutamine or G l n, C G U = Arginine or A r g, C G C = Arginine or A r g, C G A = Arginine or A r g, C G G = Arginine or A r g, A U U = Isoleucine or I l e, A U C = Isoleucine or I l e, A U A = Isoleucine or I l e, A U G = Methionine or M e t, A C U = Threonine or T h r, A C C = Threonine or T h r, A C A = Threonine or T h r, A C G = Threonine or T h r, A A U = Asparagine or A s n, A A C = Asparagine or A s n, A A A = Lysine or L y s, A A G = Lysine or L y s, A G U = Serine or S e r, A G C = Serine or S e r, A G A = Arginine or A r g, A G G = Arginine or A r g, G U U = Valine or V a l, G U C = Valine or V a l, G U A = Valine or V a l, G U G = Valine or V a l, G C U = Alanine or A l a, G C C = Alanine or A l a, G C A = Alanine or A l a, G C G = Alanine or A l a, G A U = Aspartic acid or A s p, G A C = Aspartic acid or A s p, G A A = Glutamic acid or G l u, G A G = Glutamic acid or G l u, G G U = Glycine or G l y, G G C = Glycine or G l y, G G A = Glycine or G l y, G G G = Glycine or G l y.

When using this chart, remember that nucleic acids are read from 5’ to 3’. Therefore, the first nucleotide, or letter, will always be the one closest to the 5’ end and the third nucleotide will always be the one closest to the 3’ end.

# Lecture 4 - Inheritance and Genetics

# Lecture 4.1 - Genes to Proteins

# Types of Mutations

point mutation is a change in a single nucleotide base.

Point mutations may result in…
- a missense mutation, in which the changed nucleotide base results in a single amino acid change in the protein product.
- a silent mutation, in which the changed nucleotide base does not result in any changes to the amino acids in the protein product.
- a nonsense mutation, in which the codon that includes changed nucleotdie base changes from coding for an amino acid to a stop codon that terminates translation.

An insertion is the addition of an extra nucleotide(s) within the sequence. Similarly, a deletion is the elimination of a nucleotide(s) from the sequence.

Insertions and deltions frequently result in frameshift mutations, by which the extra or missing nucleotide bases change the reading frame (the grouping of three adjacent nucleotides into codons), thus resulting in a change in the amino acids that are encoded by that nucleotide sequence.

# Lecture 4.2 - Allele Segregation

# Alleles

An allele is a version of a gene that determines which trait will be expressed. This occurs because each allele encodes slightly different versions of the same protein product.

# Segregation of Alleles

In mitosis, each daughter cell will end up with the same alleles as the original mother cell.
Therefore, if the mother cell possesses alleles B and b for gene 1, then both daughters will have both B AND b.

However, in meiosis, each gamete produced will have only one allele of each gene from the mother cell.
Therefore, if the mother cell possesses alleles D and d for gene 2, then each gamete will have either D OR d.
Which allele a gamete will inherit for one gene is independent of which allele it will inherit for another gene. This means that all of the allelic combinations shown in the diagram below are equally likely to arise in the gametes from the mother cell.

This diagram depicts a cell that undergoes meiosis. The cell has two pairs of homologous chromosomes, each with one gene, undergoing meiosis. Each gene has two alleles. One chromosome carries a gene with alleles big B or small b. The other chromosome carries a gene with alleles big H or small h. The parent cell is diploid and has a genotype of big B small b and big H small h. First, there is a cell, represented as a circle, that has two chromosomes. These chromosomes are each represented as a line in the cell. One chromosome is labelled A 1 and the other is labeled A 2. Then, an arrow labelled D N A replication points from this first cell diagram to a second cell diagram. This second circle contains four lines that represent the two copies of the two original chromosomes. The two lines labelled A 1 are paired and the other two lines labelled A 2 are paired. There are two arrows that stem from the second cell diagram, and each one points to a new cell to indicate that the original cell has divide into two new cells. In this step, known as Meiosis I, the homologous chromosomes separate. Therefore, one of these new cells contains the two paired A 1 chromosomes and the other new cell contains the two paired A 2 chromosomes. The next step is labelled Meiosis II and is when sister chromatids separate. Each of the two cells produced from Meiosis I divides to produce two new cells. Each of these new cells contains one copy of one chromosome. Therefore, a total of four cells are produced. Two of these product cells have a single copy of the A 1 chromosome, and two of these product cells have a single copy of the A 2 chromosome. Thus, the products of meiosis are 4 haploid gametes.

# Lecture 4.3 - Punnett Squares

# A List of Genetic Terminology

*NOTE: this is the genetic definition of a gene. In terms of molecular biology, a gene is defined as the DNA instructions for a product.

# Dominant v. Recessive

When in combination, different traits can be either dominant or recessive. If a trait is dominant, it will be expressed whenever an allele for that trait is present. If a trait is recessive, it will be expressed ONLY when the allele for that trait is present AND dominant alleles are absent.

For example, for a gene determines color, there may be an allele (A) that codes for red color and an allele (a) that codes for green color.
If A is dominant to a (meaning that a is recessive to A), then AA = red, Aa = red, and aa = green.

However, some traits have more complicated expression patterns. With the same example gene, incomplete dominance would result in a new, blended trait when different alleles are present.
For example: Aa = orange.

There is also codominance in which the traits of both alleles are expressed, leading to a mixed trait.
For example: Aa = red and green spots.

# Punnett Squares

Punnett Squares are a method to visually represent the likelihood of potential genotypes (and their resulting phenotypes) of the progeny of a cross (the F1 generation) based on the genotypes of the parents.

This image shows a Punnett Square, which is a 3 by 3 table. Here, the big B allele is dominant to small b, and small b is recessive to big B. Big B is associated with a green phenotype, and small b is associated with a yellow phenotype. Above the Punnett Square, the information about the genetic cross is given as follows. The cross is between two individuals that are each from parental strains. One individual is referred to as P1 and the other is P2. Both P1 and P2 have a green phenotype. The genotype of P1 is large B small b. The genotype of P2 is also large B small b. Now, in the table, each allele of P1&rsquo;s genotype is written in a column of the first row and each allele of P2&rsquo;s genotype is written in a row of the first column. These boxes represent the possible gametes of each parent. The box corresponding to the first row and first column is left blank. Thus, the second column of the first row contains a large B and the third column of the first row contains a small b. The second row of the first column contains a large B and the third row of the first column contains a small b. The remaining 4 boxes are the potential genotypes of the F1 offspring. The contents of first row are propagated down to the rows underneath. Similarly, the contents of the first column are propagated to the right to the subsequent columns. Therefore, the remaining boxes of the Punnett Square are as follows. In the second row, the second column has a genotype of large B large B and has a green phenotype. The third column of the second row has a genotype of large B small b and has a green phenotype. In the third row, the second column has a genotype of large B small b and has a green phenotype. The third column of the third row has a genotype of small b small b and has a yellow phenotype. The phenotypic ratio of offspring produced from this cross will 3 green offspring to every 1 yellow offspring.

# Lecture 4.4 - Pedigrees

# Pedigrees and Modes of Inheritance

pedigree is a visual diagram used to map out the inheritance of genetic traits within a family. Pedigrees are often useful for determining the mode of inheritance, or inheritance pattern, of a particular trait.

Genetic traits can display various modes of inheritance. They can be either dominant or recessive, and they can be either autosomal or X-linked. This course focuses on the following modes of inheritance:

# Pedigree Nomenclature

# Lecture 5 - Building with DNA

# Lecture 5.1 - Restriction Digests

# Restriction Endonucleases

restriction endonuclease, sometimes referred to simply as a restriction enzyme, cuts double-stranded DNA at palindrome sequences. Each restriction endonuclease cuts at one specific sequence. After being cut, the double-stranded DNA is left with “sticky ends”. Different restriction endonucleases leave different types of sticky ends, which are reviewed below.

# Types of Sticky Ends

Sma1 leaves blunt ends:
This image shows the sequences of two complementary strands of D N A. Each strand has a red arrow that points between two of the bases to show how and where each D N A strand is cut by the S m a l 1 enzyme. The top strand is written as 5 prime C C C arrow G G G 3 prime. The bottom strand is written as 3 prime G G G arrow C C C 5 prime. As a result of the enzyme cutting, the following 2 double-stranded D N A products are shown. The first product has a top strand of 5 prime C C C 3 prime and a bottom strand of 3 prime G G G 5 prime. The second product has a top strand of 5 prime G G G 3 prime and a bottoms strand of 3 prime C C C 5 prime.

EcoR1 leaves sticky (5’ overhang) ends:
This image shows the sequences of two complementary strands of D N A. Each strand has a red arrow that points between two of the bases to show how and where each D N A strand is cut by the Eco R 1 enzyme to produce two L-shaped pieces. The top strand is written as 5 prime G arrow A A T T C 3 prime. The bottom strand is written as 3 prime C T T A A arrow G 5 prime. As a result of the enzyme cutting, the following 2 double-stranded D N A products are shown. The first product has a top strand of 5 prime G 3 prime and a bottom strand of 3 prime C T T A A 5 prime. The second product has a top strand of 5 prime A A T T C 3 prime and a bottoms strand of 3 prime G 5 prime.

PstI leaves sticky (3’ overhang) ends:
This image shows the sequences of two complementary strands of D N A. Each strand has a red arrow that points between two of the bases to show how and where each D N A strand is cut by the P s t 1 enzyme to produce two L-shaped pieces. The top strand is written as 5 prime C T G C A arrow G 3 prime. The bottom strand is written as 3 prime G arrow A C G T C 5 prime. As a result of the enzyme cutting, the following 2 double-stranded D N A products are shown. The first product has a top strand of 5 prime C T G C A 3 prime and a bottom strand of 3 prime G 5 prime. The second product has a top strand of 5 prime G 3 prime and a bottoms strand of 3 prime A C G T C 5 prime.

# Lecture 5.2 - Compatible Ends

# Big Picture: Cloning

Cloning is the process of isolating a gene of interest (GOI) and expressing it in an organism.

vector is any carrier of DNA that can deliver that DNA, i.e. the GOI, to the cell and be replicated. One common example of a vector is a plasmid, which is a piece of small, circular DNA.

The cloning process is depicted in detail below.

# Restriction Endonucleases: Compatible Ends

Restriction endonucleases, discussed in the previous section, are used to cut out and isolate the GOI (step 3 in the diagram above) and to cut the vector so that the GOI can be integrated into the vector (step 4 in the diagram above).
In order for the GOI to integrate into the vector, the two need to be cut in a way that gives them compatible ends. In this way, they can match up and ligate into a new circular plasmid without any breaks. The different types of compatible ends are detailed below.

Note that the GOI and the vector do not need to be cut with the same enzyme to be compatible. In the case of blunt ends, all blunt ends can match with other blunt ends. In the case of sticky ends, only the overhanging base pairs need to match up.
However, ligating DNAs from two different restriction sites may prevent the ligation point from serving as a restriction site in the future because you alter the sequence on the other half of the restriction site (in the case of blunt ends) or around the overhanging base pairs (in the case of sticky ends).

# Lecture 5.3 - PCR

# The PCR Process

The following diagram depicts the process of PCR (Polymerase Chain Reaction), which has 3 main steps:

  1. denaturation
  2. primer annealing
  3. extension (synthesis of new DNA by the DNA polymerase, Taq Polymerase)

This diagram illustrates the process of P C R described above. The process starts with 1 molecule of double stranded D N A, which is shown as a pair of lines. Each line has a 5 prime or 3 prime label at each end. The top line is black and and the complementary sequence of the bottom strand is gray. An arrow labelled cycle 1 feeds into step 1, which is annotated Denature at 95 degrees Celsius. The subsequent illustration shows that the previously paired D N A strands are now separated. Step 2 is annotated Anneal primers at 55 degrees Celsius. The corresponding illustration shows that smaller D N A fragments, or primers, have paired with the orignal D N A strands. The 5 prime end of each primer is paired at the 3 prime end of the D N A strand. The top black D N A strand is paired with a gray primer. The gray primer follows the same complementary sequence as the bottom, gray D N A strand. Similarly, the bottom gray D N A strand is paired with a black primer, which follows the sequence of the top, black strand. Step 3 is annotated Extend D N A at 65 degrees Celsius. The corresponding illustration shows that each of the primers has been extended to the full length of the D N A strand that it paired with. Therefore, the end products are 2 double stranded D N A molecules, where each is composed of one black and one gray strand. This molecule is the same as the starting double stranded D N A molecule. Next, there is an arrow that circles from the prodcuts of the cycle back up to step 1. This arrow is labelled cycle 2 and indicates that steps 1 through 3 will be repeated. After cycle 2, there will be 4 molecules of double stranded D N A. Another circling arrow is labelled cycle 3, which will result in 8 molecules of double stranded D N A. A final, dashed arrow indicates that after 30 cycles, there will be greater than 10 to the ninth double stranded D N A molecules.

It is important to note that this reaction, while based off of the principle of DNA replication, happens in a test tube (not a cell!) with the goal of amplifying a DNA sequence.

# Notes about PCR Primers

The primers used in this reaction are DNA primers, not RNA primers, because they are synthetically made and added to the test tube. The primers should anneal to the outermost region of the sequence that you would like to amplify. For each reaction, you should have two types of primers: one to extend along the top strand of the DNA sequence and one to extend along the bottom strand.