Textbook: Urry et al. (2022). Campbell Biology, Australian and New Zealand Version (12th edition). Pearson. Australia. Chapter 14.
Learning outcomes:
Topic:
Examine the nature and flow of genetic information.
Predict outcomes of simple Mendelian inheritance.
Explain how complex organisms develop from a single cell.
Lecture:
Understand and apply Mendel’s law of segregation.
Understand and apply Mendel’s law of independent assortment.
Use probability laws to determine phenotypic ratios.
Gene Definition
Definition of a gene.
Terminology
Allele: An alternate form of a gene.
Homozygous
Heterozygous
Gamete
True-breeding/pure-breeding
Self-cross
Mendel’s First Law: The Law of Segregation
The two alleles for a heritable character segregate during gamete formation and end up in different gametes.
Mendel’s Second Law: The Law of Independent Assortment
Two or more genes assort independently.
Each pair of alleles segregates independently of any other pair of alleles during gamete formation.
Independent assortment leads to genetic variation.
Problem
How many unique gametes could be produced through independent assortment by a diploid individual with the genotype AaBb?
Mendelian Genetics Part 2
Law of Segregation in Action
Monohybrid cross.
P Generation: True-breeding yellow (YY) x True-breeding green (yy).
Yellow (Y):
Genotype: YY
Gametes: Y
Green (y):
Genotype: yy
Gametes: y
F1 Generation: Yy (all yellow-seeded).
Phenotype: Yellow
Genotype: Yy
Gametes: Y, y
Law of Segregation in Action
F2 Generation: Self-cross of F1 (Yy x Yy).
Law of Independent Assortment in Action
Dihybrid cross.
Two genes:
Yellow gene:
Two alleles: Y > y
Yellow > green
Round gene:
Two alleles: R > r
Round > wrinkled
P Generation: True-breeding round yellow (YYRR) x True-breeding green wrinkled (yyrr).
Yellow Round: YYRR
Gametes: YR
Green Wrinkled: yyrr
Gametes: yr
F1 Generation: YyRr (all round yellow).
Law of Independent Assortment in Action
F2 cross: self-cross of F1 generation.
Gametes from F1: YR, Yr, yR, yr.
Phenotypic ratio in F2: 9:3:3:1.
Summary: Mendelian Phenotypic Ratios
Monohybrid cross:
One gene: 3:1.
Dihybrid cross:
Two genes: 9:3:3:1.
Test Cross
A test cross can be used to determine genotype.
Question: Is the purple plant homozygous (PP) or heterozygous (Pp)?
Cross plant with unknown genotype to homozygous recessive plant.
Predict a 1:1 phenotypic ratio in offspring if heterozygous.
Purple plant (P-) x White plant (pp).
If purple plant is PP:
F1: All offspring are purple (Pp).
If purple plant is Pp:
F1: 1/2 offspring are purple (Pp), 1/2 offspring are white (pp).
Summary: Mendelian Phenotypic Ratios
Monohybrid cross:
One gene, 3:1.
Dihybrid cross:
Two genes, 9:3:3:1.
Test cross:
One gene, expect a 1:1 ratio.
Two genes, expect a 1:1:1:1 ratio.
Problem
Mary is a farmer breeding pea plants. Seed colour in pea plants is controlled by a single gene. Mary crosses true breeding yellow-seeded and green-seeded pea plants. The F1 generation are all yellow-seeded. She then allows the F1 yellow-seeded offspring to self-pollinate to produce the F2 generation. There are 8,500 pea plants in the F2 generation. What are the expected numbers of yellow-seeded and green-seeded pea plants in the F2 generation?
Mendelian Genetics Part 3
Learning outcomes:
Topic:
Examine the nature and flow of genetic information.
Predict outcomes of simple Mendelian inheritance.
Explain how complex organisms develop from a single cell.
Lecture:
Understand and apply Mendel’s law of segregation.
Understand and apply Mendel’s law of independent assortment.
Use probability laws to determine phenotypic ratios.
Probability Laws
Simple probability.
Multiplication rule.
Addition rule.
Multiplication (Product) rule:
Probability of two or more independent events occurring together is the product of the probabilities that each event will occur by itself.
Addition (Sum) rule:
Probability of either of two such mutually exclusive events occurring is the sum of their individual probabilities.
Problem
Consider a yellow round pea of the following genotype, YyRr.
What is the probability that this plant will produce a YR gamete? 1/4
If this plant is self-crossed, what is the probability of a YYRR genotype in the next generation? 1/16
If the plant is self-crossed, what is the probability of a YYRR or a yyrr genotype in the next generation? 1/16 + 1/16 = 2/16 = 1/8
Problem
Two true-breeding stocks of plants were crossed. One parent had red, axial flowers and the other had white, terminal flowers. All offspring produced during the F1 generation had red, axial flowers. 1,000 F2 generation offspring resulted from the self-cross of F1 individuals. Approximately, how many of the F2 generation would you expect to have red, terminal flowers? Assume independent assortment. Round your answer to the nearest whole number.
Red, terminal flowers = 3/16.
(3/16) * 1000 = 187.5
Approximately 188.
Biology with Trixie
Coat colour in Labradors is controlled by two genes, B and E.
Individuals with CF have two mutated copies of gene.
Phenotype:
Chronic obstructive lung disease.
Exocrine pancreatic insufficiency.
Elevated sweat chloride concentration.
Infertility.
Autosomal Dominant Inheritance: Achondroplasia
Form of dwarfism caused by dominant allele.
Individuals with Achondroplasia have one mutated copy of FGFR3 gene.
Two copies of dominant mutated allele is lethal.
Multiple Choice Question
Michael has Achondroplasia. His partner is of typical stature. What is the probability that they will have a biological male child of typical stature?
1/2 (correct)
1/4
1/8
1/16
Extension to Mendelian Genetics Part 3
Learning outcomes:
Topic learning outcomes:
Examine the nature and flow of genetic information
Predict outcomes of simple Mendelian inheritance
Explain how complex organisms develop from a single cell
Lecture learning outcomes:
Describe and solve complex inheritance patterns
Use a pedigree to depict autosomal dominant and autosomal recessive inheritance of a trait
Compare and contrast the human X-Y system to other systems
Use a pedigree to depict X-linked dominant and X-linked recessive inheritance of a trait
Understand and describe X- inactivation
Thomas Hunt Morgan
American Scientist.
Used Drosophila melanogaster to study genetics.
1910 discovered first mutation in Drosophila.
White eyed mutant.
Established the chromosomal theory of inheritance.
Awarded Nobel Prize of Medicine in 1933.
Genes on X Chromosome Show Unique Inheritance Patterns
w+ allele = wild type, red eyes.
w allele = mutant, white eyes.
Males have white eyes when carry one copy of w allele.
Females must have two copies of w to have white eyes.
Gene for eye colour on X chromosome
Inheritance of an X-Linked Recessive Trait: Red-Green Colour Blindness
Can not distinguish shades of red and green.
Caused by mutation in OPN1LW or OPN1MW genes.
More common in males.
Inheritance of a X-Linked Dominant Trait: Alport Syndrome
Rare disorder.
Phenotype includes kidney disease, hearing loss, eye abnormalities.
Affected people have a mutation in the COL4A5 gene.
More common in biological females (often lethal in males)
How to Determine the Mode of Inheritance in a Pedigree
Are there affected individuals in every generation?
Do affected parents have affected offspring?
Do affected males have affected sons?
What is the ratio of affected males to females?
Rare: Most likely X-linked recessive.
Most likely Autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive
X-Inactivation
X inactivation occurs in biological female mammals.
Biological female mammals (XX):
One X chromosome is inactivated in each cell during early embryonic development.
Inactive X condenses = Barr body.
Most of genes in Barr body not expressed.
In the ovaries, Barr body chromosomes reactivated.
Multiple Choice Question
Cinnabar eye colour is an X-linked, recessive characteristic in fruit flies. If a female having cinnabar eyes is crossed with a male having wild type, red eyes, what percent of the F1 males will have cinnabar eyes?
0%
25%
50% (correct)
75%
100%
Biology with Trixie
What can the dogs of Chernobyl teach us about living for decades under extreme radiation?