Genetics -. Basic Principles of Mendelian Genetics and Patterns of Inheritance -Molecular Genetics & Inheritance -. Protein Synthesis - Mutations -. Manipulation of DNA -. ABO blood groups and Rh Factors Evolution - Theories on the origin of life on Earth -. Theories of Evolution

1. Course Title: General Biology
(Biol. 1012)
Chapter 4. Genetics and Evolution
1

2. outline
4. GENETICS & EVOLUTION
4.1 Genetics
4.1.1. Basic Principles of Mendelian Genetics and
Patterns of Inheritance
4.1.2. Molecular Genetics & Inheritance
4.1.3. Protein Synthesis
4.1.4. Mutations
4.1.5. Manipulation of DNA
4.1.6. ABO blood groups and Rh Factors
4.2. Evolution
4.2.1. Theories on the origin of life on Earth
4.2.2. Theories of Evolution
2

3. Introduction to Genetics
Genetics is the scientific study of heredity &
variation of organisms
Heredity deals with
→ how traits (such as hair color, eye color, height, and
even genetic disorders) pass from parent to offspring.
→ The physical and molecular nature of hereditary
material.
→ How the genetic material store information and express
itself.
3

4. Genetics terms you need to know:
Alleles – two genes that occupy the same position on homologous
chromosomes and that cover the same trait (like ‘flavors’ of a trait).
Locus – a fixed location on a strand of DNA where a gene or one of
its alleles is located.
Character: A heritable feature that varies among individuals, such
as flower color.
Trait: Each variant for a character, such as purple or white flowers
Phenotype – the physical appearance of an organism i.e. the
physical characteristic (what it looks like) ex: tall or short
Genotype – the genetic makeup of an organisms
Gene – a unit of heredity; a section of DNA
sequence encoding a single protein
Genome – the entire set of genes in an organism
4

5. Homozygous – having identical genes/ alleles (one from
each parent) for a particular characteristic(either both
dominant, TT, or both recessive, tt). it also known as
purebred
Heterozygous – having two different genes/ alleles for a
particular characteristic (one dominant, one recessive: Tt). it
also known as hybrid
Dominant – the allele of a gene that masks or suppresses
the expression of an alternate allele; the trait appears in the
heterozygous condition.
Recessive – an allele that is masked by a dominant allele;
does not appear in the heterozygous condition, only in
homozygous. 5

6. Hybridization: Mating, or crossing, of two varieties
Monohybrid Cross: a genetic cross involving a single pair of
genes (one trait); parents differ by a single trait.
P = Parental generation
F1 = First filial generation; offspring from a genetic cross.
F2 = Second filial generation of a genetic cross
Dihybrid Cross: Cross of two different true-breeding strains
(homozygotes) that differ in two traits.
Nomenclature = Representing characters using symbols
= most of the time, we use letters.
A. Dominant Genotype & Wild Type Phenotype – We use
capital letter or a+ or + etc
B. Recessive Genotype & Mutant Phenotype – We use
Lowercase letters or a- or – etc 6

7. Gregor Mendel
Austrian monk born in 1822.
Attended the University of Vienna
Relied upon a background of
mathematics, physics, & chemistry.
He established his Principles of
Inheritance by breeding garden peas, &
laid the foundation of the science of
genetics.
 He was the 1st to quantify the result of
his expt.
 The Father of Genetics
7

8. Mendel’s Hybridization Expt
He study heredity on peas (1856 – 1863).
His test subject : Garden peas (Pisum sativum) because:
1. Easy to cultivate.
2. Naturally self-pollinating plant but it is easy for
artificial cross-pollination.
3. Having characters which show pair of contrasting
traits (no intermediates).
8
He carefully focus on 7 characters
(14 varieties or contrasting traits).
→ Ex. Stem length (Tall ₰ Short).

9. Character Traits
Dominant Recessive
Flower color
Purple White
Flower position
Axial Terminal
Seed color
Yellow Green
Seed shape
Round Wrinkled
Pod shape
Inflated Constricted
Pod color
Green Yellow
Stem length
Tall Dwarf

10. For each contrasting traits, conduct his expt in 3 stages.
1. Test the purity of each variety by letting them self
pollinate for a no of generation.
→ True or Pure breed = A plant, that when self-fertilized,
only produces offspring with the same traits.
2. Cross- pollinate plants with contrasting traits and see
what the offspring (F1) look like.
→ His actual expt starts
→ By Artificial Cross- pollinate
 To prevent self pollination, he cut away the pollen-bearing
male parts ₰ then dusting pollen from another plant onto
the flower.
3. F1 generation (hybrid) plant allowed it to self pollinate
and see the result (F2)
10

11. Monohybrid Cross
• Parents differ by a single trait.
• Crossing two pea plants that differ in stem size,
one tall one short
T = allele for Tall
t = allele for dwarf
TT = homozygous tall plant
t t = homozygous dwarf plant
T T  t t 11

12. Monohybrid cross for stem length:
T T  t t
(tall) (dwarf)
P = parentals
true breeding,
homozygous plants:
F1 generation
is heterozygous:
T t
(all tall plants)
12

13. Finding the F1 generation Using a Punnett Square
1. Determine the genotypes of the parent organisms
2. Write down your "cross" (mating) and draw a p-square
3. "Split" the letters of the genotype for each parent & put them
"outside" the p-square
4. Determine the possible genotypes of the offspring by filling in the p-
square
5. Summarize results (genotypes & phenotypes of offspring)
T t T t
T t T t
T T
t
t
Genotypes:
100% T t
Phenotypes:
100% Tall plants
Parent genotypes:
TT and t t
Cross
T T  t t
13

14. Monohybrid cross: F2 generation
• If you let the F1 generation self-fertilize, the next
monohybrid cross would be:
T t  T t
(tall) (tall)
T T T t
T t t t
T t
T
t
Genotypes:
1 TT= Tall
2 Tt = Tall
1 tt = dwarf
Genotypic ratio= 1:2:1
Phenotype:
3 Tall
1 dwarf
Phenotypic ratio= 3:1
14

15. The Experiment
P generation
(true-breeding
parents)
F1 generation
F2 generation
of plants
have purple flowers
of plants
have white flowers
Purple
flowers
White
flowers
All plants have
purple flowers
Fertilization
among F1 plants
(F1  F1)

3
4
1
4
Another
example:
Flower
color

16. Character Traits
Dominant Recessive
Flower color
Purple White
Flower position
Axial Terminal
Seed color
Yellow Green
Seed shape
Round Wrinkled
Pod shape
Inflated Constricted
Pod color
Green Yellow
Stem length
Tall Dwarf
Ratio
3.15:1
3.14:1
3.01:1
2.96:1
2.95:1
2.82:1
2.84:1
3:1
NOTE
Mendel
observe the
same
Patterns of
Inheritance
for all 7
characters
of pea
plant.
16

17. Mendel’s Conclusions
1. Heredity are determined by factors (genes) that are
passed from one generation to the next.
2. Each character is determined by a pair of factors.
→ Each trait is controlled by one gene occurring in two
contrasting forms – the different forms of each gene
are called alleles
 Example, the gene for plant height has a tall allele and a
short allele.
→ Factors / alleles found in pairs.
→ For each trait, an organism inherits 2 alleles (pair of
factors), one from each parent.
17

18. 3. Law of Dominance-
 When the 2 factors of a pair are different, only one is fully
expressed/ shows in a hybrid individual
→ This can be determined by w/h allele shows in a hybrid
individual.
→ The expressed one is dominant allele and the masked one
is recessive allele
 Example
→ A cross b/n a pure tall & a pure short will give 100% tall
offspring.
→ Therefore, the tall allele is dominant & the short allele is
recessive.
→ Recessive alleles will be shown ONLY if no dominant allele is
present.
18

19. 5. Law of Independent Assortment
During gamete formation, factors belonging to
different pairs assort independently of one
another.
“Members of one gene pair segregate
independently from other gene pairs during
gamete formation”
Genes get shuffled – these many combinations
are one of the advantages of sexual
reproduction
19
4. Law of Segregation
Alleles (found in homologous pairs) are segregated (separated)
from each other during meiosis so that each gamete (egg, sperm)
carries only a single copy of each gene
6. During fertilization, factors of a pair will recombine so that factors /
alleles are always found in pairs.
New pairs form during fertilization

20. Dihybrid crosses
• Matings that involve parents that differ in two
genes (two independent traits)
For example, flower color:
P = purple (dominant)
p = white (recessive)
and stem length:
T = tall t = short
20

21. Dihybrid cross: flower color and
stem length
TT PP  tt pp
(tall, purple) (short, white)
Possible Gametes for parents
T P and t p
F1 Generation: All tall, purple flowers (Tt Pp)
TtPp TtPp TtPp TtPp
TtPp TtPp TtPp TtPp
TtPp TtPp TtPp TtPp
TtPp TtPp TtPp TtPp
tp tp tp tp
TP
TP
TP
TP
21

22. Dihybrid cross F2
If F1 generation is allowed to self pollinate,
Mendel observed 4 phenotypes:
Tt Pp  Tt Pp
(tall, purple) (tall, purple)
Possible gametes:
TP Tp tP tp
Four phenotypes observed
Tall, purple (9); Tall, white (3); Short, purple (3); Short white (1)
TTPP TTPp TtPP TtPp
TTPp TTpp TtPp Ttpp
TtPP TtPp ttPP ttPp
TtPp Ttpp ttPp ttpp
TP Tp tP tp
TP
Tp
tP
tp
22

23. Dihybrid Cross
9 Tall purple
3 Tall white
3 Short purple
1 Short white
TTPP TTPp TtPP TtPp
TTPp TTpp TtPp Ttpp
TtPP TtPp ttPP ttPp
TtPp Ttpp ttPp ttpp
TP Tp tP tp
TP
Tp
tP
tp
Phenotype Ratio = 9:3:3:1
23

24. Genotype Ratios (9): Four Phenotypes:
1 TTPP
2 TTPp
2 TtPP
4 TtPp
1 TTpp
2 Ttpp
1 ttPP
2 ttPp
1 ttpp
Dihybrid Cross: 9 Genotypes
Tall, purple (9)
Tall, white (3)
Short, purple (3)
Short, white (1)
24

25. 1. Monohybrid Cross
F2 result
Genotypic ratio = 1 : 2 : 1
Phenotypic Ratio = 3 : 1
Genotypic Class = 3
Phenotypic Class = 2
Crosses involving 1 & 2 levels of hybridity.
Backcross / Testcross
Genotypic ratio = 1:1
Phenotypic Ratio = 1:1
 Genotypic Class=2
Phenotypic Class=2
2. Dihybrid Cross
F2 result
Genotypic ratio =
1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1
Phenotypic Ratio = 9 : 3 : 3 : 1
Genotypic Class = 9
Phenotypic Class = 4
Backcross / Testcross
Genotypic ratio = 1:1:1:1
Phenotypic Ratio=1:1:1:1
 Genotypic Class=4
Phenotypic Class=4 25

26. Beyond Dominant & Recessive Alleles
There are some exceptions to Mendel’s important
principles.
Some alleles are neither dominant nor recessive, and some
are controlled by multiple alleles or many genes.
Examples of Non-Mendelian Inheritance Patterns
 Incomplete Dominance
 Codominance
 Multiple Alleles
 Genetics –Environment Interaction
26

27. Incomplete Dominance
27
A case in which one allele is not completely dominant over
another is called incomplete dominance.
a new phenotype appears in the heterozygous condition as a
BLEND of the dominant and recessive phenotypes.
Ex - Snapdragon flowers
→Dominant Red (RR) + Recessive White (rr) = Hybrid Pink (Rr)
RR = red rr = white Rr = pink

28. Co-dominance
28
When both alleles contribute to the phenotype, we call that
co-dominance.
Colors are not blended; they appear separately.
Examples: Roan cattle
 Cattle can be
→ Red (RR)--- all red hairs
→ White (WW)---- all white hairs
→ Roan (RW)---- red & white hairs together

29. Multiple Alleles
29
Genes that have more than two alleles are said to have
multiple alleles.
This means that more than two possible alleles exist in a
population.
→However, only two alleles are inherited.
Example: ABO Blood Types
→ ABO Blood types can be determined by 3 alleles (IA ,IB & i)
→ Allele IA & allele IB are Co-dominant to each other but both
are Complete Dominant over allele i

30. Phenotype
Possible
Genotype(s)
Allele
(antigen)
on RBC
surface
Can
Donate
Blood
To
Can
Receive
Blood
From
A
IAi
IAIA A A, AB A, O
B
IBi
IBIB B B, AB B, O
AB IAIB AB AB
A, B,
AB, O
O ii O
A, B,
AB, O O 30

31. Genetics –Environment Interaction
31
Characteristics are not solely determined by genes, but they are
also determined by the interaction b/n genes & the environment
Examples of influence of environment on phenotype..
→Age of onset (pattern baldness)
→Sex (milk production, horn formation)
 Male Pattern Baldness
– Autosomal trait influenced by sex
hormones & age: onset after 30 years old.
– Dominant in males & Recessive in
females
• B_ = bald in males;
• bb = bald in females

32. Molecular Genetics
The transmission of hereditary information took place in the
nucleus, more specifically in structures called chromosomes.
The hereditary information was thought to reside in genes
within the chromosomes.
Chemical analysis of nuclei showed chromosomes are made
up largely of proteins called histones and nucleic acids.
32
Chromosome consists of :-
1. Nucleic Acid
 DNA or RNA
Genetic Material
2. Protein Coat (Histone)

33. DNA
chromatin
chromatin fibers
fibers connected to
chromosome scaffold
Condenced scaffold
Chromosome
33

34. Nucleic Acids are composed of :-
1. 5-carbon (Pentose) Sugar
- Deoxyribose (in DNA)
- Ribose (in RNA)
2. Nitrogenous Base
i. Purines (2 Rings)
- Adenine (A)
- Guanine (G)
ii. Pyrimidines (1 Ring)
- Cytosine (C)
- Thymine (T)
- Uracil (U)
3. Phosphate group attached
to 5’ carbon
Structure of DNA:
34

35. Nucleic Acids are formed as follow:-
→ 1st Base + Sugar Nucleoside
→ 2nd Nucleoside + Phosphate Nucleotide.
→ 3rd Nucleotide + Nucleotide + … Polynucleotide
(= Nucleic Acid)
35
Nucleotide
→ Smallest Unit of Genetic Material
→ Monomer or Repeating Unit or
Building Unit of Nucleic Acid
(DNA or RNA )

36. Notes on Structure of DNA:
1. Relative Proportions (%) of Bases in DNA.
Erwin Chargaff (1950)
Discovered a 1:1 ratio of adenine to thymine and
guanine to cytosine in DNA samples from a variety
of organisms.
Chargaff’s Rules
36
A = T
C = G
A + C = T + G
Purine = Pyrimidine

37. 37
2. The structure of DNA was determined by James Watson
and Francis Crick (1953).
 According to Watson & Crick DNA Model:
Backbone made of alternating phosphate and sugar molecules (linked
together by Phosphodiester Bond)
Two strands twist to form a Double Helix
Two strands of a DNA are Complementary (not identical).
Complementary base pairs from opposite strands are bound together
by weak hydrogen bonds.
→ “A” pairs with “T” (2 H-bonds), &
→ “G” pairs with “C” (3 H-bonds).
5’-TATTCCGA-3’
3’-ATAAGGCT-5’
 Two DNA strands are Anti-parallel (Opposite directions)

38. DNA Double Helix
16-38

39. 5’ end
3’ end
39

40. DNA Replication
DNA Replication occurs during chromosome duplication;
an exact copy of the DNA is produced with the aid of DNA
Polymerase
The process is Semi-Conservative - Each new double
helix is composed of old (parental) strand and new
(daughter) strand
During DNA replication, base pairing enables existing DNA
strands to serve as templates fornewcomplementarystrands
Parental DNA
DNA Template
New DNA
40

41. DNA
Polymerase III
 Build daughter DNA strand
from complementary parent
template (pattern)
 add new complementary bases
 DNA polymerase III main base
layer
41

42. 1. Unwinding-
→ Two strands unwind & unzip (weak H-bonds break);
→ Helicase helps molecule unwind
2. Complementary base pairing –
→ Complementary nucleotides available in cell, pair with those
of old strand with help of DNA Polymerase III
3. Joining-
→ The new strand is joined to form a daughter strand by
ligase, now have two daughter strands
4. Editing & Proofreading DNA.
→ Correcting Replication Errors by the help of DNA
polymerase I
 Basic principle of DNA replication is quite simple; however actual
process involves complex biochemical mechanisms w/h requires
more than a dozen enzymes and other proteins. 42
Steps to DNA Replication

43. Structure of RNA
 Single Stranded nucleotide
 Each nucleotide of RNA
contains:
– Ribose Sugar
– Phosphate
– Nitrogen base (A, G, C, U)
 There are 3 Classes of RNA
that participate & play d/t
roles In protein synthesis.
1. Messenger RNA
2. Ribosomal RNA
3. Transferal RNA 43

44. 1. Messenger RNA (mRNA)
44
Carries copy of instruction for proteins from DNA
to the ribosomes.
Contains information (messages/codon) copied the
gene that is translated in to protein.
2. Ribosomal RNA (rRNA)
rRNA is a structural component
of the ribosome
Important for protein syntesis.

45. 3. Transferal RNA (tRNA)
45
Are molecules carry amino
acids to the ribosome for
incorporation into a polypeptide.
Transfers amino acids to the
respective codon on the mRNA
during protein synthesis.
 Aminoacyl-tRNA synthetases
add amino acids to the acceptor
arm of tRNA
 Anticodon loop contains 3
nucleotides complementary to
mRNA codons

46. Genetic code: language that relates the series of nucletides in
mRNA to the amino acids specified.
The sequence of nucleotides in the mRNA determines the
amino acid order for the protein.
 Every three bases (triplet) along the mRNA makes up a codon.
 Each codon specifies a particular amino acid.
 Codons are present for all 20 amino acids.
 64 condons are possible from the triplet combination of A, G,
C, & U.
→ 61 codons code for amino acids.
→ 3 (UAA, UGA, UAG) are stop codons (or nonsense
codons) that terminate translation(have no tRNAs!).
→ 1 start codon (initiation codon), AUG, coding for
methionine.
Genetic Code
46

47. Genetic code
A
U
U C
C
A
Leu
Leu
Leu
Leu
Val
Val
Val
Val
Ser
Ser
Ser
Ser
Pro
Pro
Pro
Pro
Thr
Thr
Thr
Thr
Ala
Ala
Ala
Ala
Arg
Arg
Arg
Arg
Gly
Gly
Gly
Gly
UUU
UUC
Phe
Phe
Leu
Leu
Ile
Ile
Ile
AUU
AUC
AUA
UAU
UAC
Tyr
Tyr
CAU
CAC
His
His
Gln
Gln
AAU
AAC
Asn
Asn
Lys
Lys
Asp
Asp
Glu
Glu
Trp
Cys
Cys
Ser
Ser
Arg
Arg
UCU
UCC
UCA
UCG
G
U
C
A
G
CUU
CUC
CUA
CUG
CCU
CCC
CCA
CCG
CGU
CGC
CGA
CGG
U
C
A
G
G
ACU
ACC
ACA
ACG
U
C
A
G
U
C
A
G
GUU
GUC
GUA
GUG
GCU
GCC
GCA
GCG
GGU
GGC
GGA
GGG
Stop
5' 3'
UUA
UUG
Met*
AUG
UAA
UAG
Stop
Stop
CAA
CAG
AAA
AAG
GAU
GAC
GAA
GAG
UGA
UGG
UGU
UGC
AGU
AGC
AGA
AGG
*AUG signals translation initiation as well as coding forMet
47

48. 48
Protein Biosynthesis (Gene Expression)
1. Transcription
2. Translation

49. 49
protein
Transcription Translation
DNA
replication
DNA mRNA
Reverse transcriptase
RNA
replication
Flow of genetic information in a cell
DNA RNA Protein Phenotype
The “Central Dogma”

50. 1. Transcription
 Translation- Process of copying RNA molecule from
DNA template by RNA Polymerase (Transcriptase )
enzyme.
 Only one strand (Template Strand) of DNA act as
template
 Template Strand: strand of the DNA double helix used to
make RNA
 Coding Strand: strand of DNA that is complementary to
the template strand
 RNA Polymerase: the enzyme that synthesizes RNA from
the DNA template
50

51. Transcription proceeds through:
– Initiation – RNA polymerase identifies where to begin
– Elongation – RNA nucleotides are added to the 3’ end of
the new RNA
– Termination– RNA polymerase stops transcription when
it encounters terminators in the DNA sequence
51
 Major activities in Transcription
1. Enzymes unzip DNA
2. Free RNA nucleotides pair with complementary nucleotides on
one of the DNA strands
 Follow base pairing rules
3. When base pairing is complited, mRNA leaves nucleus & heads
towards cytoplasm (Ribosome)
4. Proteins are synthesized at ribosome

52.  The genetic information in mRNA (nucleotides) molecules
is translated into the amino acid sequences of polypeptides
according to the specifications of the genetic code.
 Synthesis of protein according to the information (codon)
contained in mRNA
 Translation proceeds through
– Initiation – mRNA, tRNA, and ribosome come together
– Elongation – tRNAs bring amino acids to the ribosome
for incorporation into the polypeptide
– Termination – ribosome encounters a stop codon &
releases polypeptide
52
2. Protein Synthesis: Translation

53. There are 3 stages in translation:
1. Initiation begins with
mRNA binding to the
ribosome.
2. Elongation proceeds as the
next tRNA molecule delivers
the next amino acid, and a
peptide bond forms between
the two amino acids.
53

54. 3. Termination: Translation continues until a stop codon
(UAA, UAG, or UGA) is reached and the completed
protein is released.
Often the first amino acid (methionine) is not needed
and it is removed after protein synthesis is complete.
54

55. • Mutations are changes in the genetic material
– it can be beneficial, deleterious, or have no effect (neutral)
• There are two main types of mutations:
1. Gene mutations & 2. Chromosomal mutations
1. Gene mutations
• Point mutations: involve changes in one or a few nucleotides;
there are three main types:
– Substitutions: one base is substituted with another
– Insertions: an additional base is inserted into the nucleotide
sequence
– Deletions: a base is removed from the nucleotide sequence
55
Mutations

56. Insertions and deletions are called frameshift mutations
because they shift the letters of the genetic message. Change
the code  different amino acids  useless proteins 
major problems!
56

57. • Chromosomal mutations involve changes in the structure or number
(e.g. trisomy) of chromosomes.
• There are four main types:
– Deletion: loss of all or part of a chromosome
– Duplication: extra copies produced
– Inversion: reverse the direction of parts of chromosomes
– Translocation: part of one chromosome breaks off and attaches to
another
57
2. Chromosomal Mutations

58. 58