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Chapter 12
Vocabulary:
Hybrids
True-breeding
F1
F2
Dominant
Recessive
Genes
Alleles
Homozygous
Heterozygous
Genotype
Phenotype
Pedigree
Polygenic
Pleiotropic
Incomplete dominance
Codominant
Epistasis
Outline
12.2 Mystery of Heredity
True-breeding
Hybrids
Gregor Mendel
Austrian monk with a math and science background
Worked with pea plants
Usually self-pollinate, but can cross-pollinate
Used true-breeding plants, reciprocal crosses and hybrid self-fertilization
Worked with discrete traits
12.2 Monohybrid Crosses: Principle of Segregation
Monohybrid cross – 1 trait
Principle of segregation – for every trait there are 2 alleles. These alleles randomly
separate and are then randomly recombined
Parents pass on genes
Offspring get one copy of each gene from each parent
Alleles are different versions of genes
Alleles are separate entities
Presence of an allele doesn’t guarantee its expression
P:
this is the genotype of the parents
Gametes: these are all the possible alleles that the parents can pass on
Cross: this is the cross of the parent gametes
F1:
this is the genotypes of all of the possible offspring
F2:
this is the product of a cross between F1 individuals
Answers should be given as ratios
Genotype:
# Homozygous dominant:
# Heterozygous:
# Homozygous recessive
Phenotype – physical expression of a trait (# expressing one phenotype: # expressing
another) (i.e. 3purple:1white)
Pedigrees of dominant traits tend to have the trait expressed in every generation
Pedigrees of recessive traits show unaffected parents having affected offspring (they are
carriers) and only 1 affected parent with unaffected offspring
12.3 Di-hybrid Crosses: Principle of Independent Assortment
Di-hybrid crosses (2 traits)
Principle of independent assortment – alleles of each gene (traits) randomly separate and
are then randomly recombined
AABB x aabb
All offspring will be AaBb
AaBb x AaBb
Use the foil method to figure out the combination of alleles that can be passed on. I
always stress the difference between genes and alleles. They have to pass on 1 allele for
each gene.
AB, Ab, aB, ab on the top and side.
Answers should again be as a ratio. i.e. 9tall purple plants: 3tall white plants: 3 dwarf
purple plants: 1 dwarf white plant
12.4Probability: Predicting the Results of Crosses
If an event is certain it has a probability of 1
If an event is impossible it has a probability of 0
Laws of Probability if free earlobes are dominant over attached earlobes:
Multiplicative law is used for genotypes – the probability of 2 or more
independent events occurring together is the product of their chance of occurring
separately.
Ee x Ee
The chance of EE is ½ x ½ = ¼
The chance of Ee is ½ x ½ = ¼
The chance of eE is ½ x ½ = ¼
The chance of ee is ½ x ½ = ¼
Additive law is used for phenotypes – the chance of an event that can occur in 2
or more independent ways is the sum of the individual chances
Chance of free earlobes is 1/4 + ¼ +1/4 = ¾ or 75%
Chance of attached earlobes is ¼ or 25%
Answer should be written as:
Genotypic ratio ¼ homozygous dominant: ½ heterozygous: ¼ homozygous recessive OR
1 homozygous dominant: 2 heterozygous: 1 homozygous recessive OR
25% homozygous dominant: 50% heterozygous: 25% homozygous recessive
Phenotypic ratio ¾ free: ¼ attached OR
3 free: 1 attached OR
75%free: 25% attached
The important thing is that there are words to go with the numbers!!!
Laws of probability for free earlobes (A) are dominant over attached (a) and widow’s
peaks (B) are dominant over flat hairlines (b). For the cross between AaBb and AaBb:
From a monohybrid cross for earlobes:
Chance of free ears is ¾
Chance of attached ears is ¼
From a monohybrid cross for hairline:
Chance of widow’s peak is ¾
Chance of flat hairline is ¼
Multiplicative law:
Chance of free ears with widow’s peak is ¾ + ¾ = 9/16
Chance of free ears with flat hairline is ¾ + ¼ = 3/16
Chance of attached ears with widow’s peak is ¼ + ¾ = 3/16
Chance of attached ears with flat hairline is ¼ + ¼ = 1/16
Answer should be written as 9 free ears with widow’s peak: 3 free ears with flat hairline:
3 with attached ears and a widow’s peak: 1 with attached ears and a flat hairline
Generally, my students are happy just to be able to successfully work the crosses and
only a select few of them would even want to learn an alternate way to get the answers so
I DON”T cover how to work crosses mathematically and I don’t expect them to be able
to perform them this way. However, if you have students (in mass or not) that prefer to
work them this way they can also get credit for this method.
12.5 The Testcross: Revealing Unknown Genotypes
If an individual shows the recessive phenotype its genotype is known (homozygous
recessive)
If an individual show the dominant phenotype its genotype is unknown (either
homozygous dominant or heterozygous)
Crossing an unknown individual with a homozygous recessive individual will allow one
to determine the genotype of the unknown.
12.6 Extensions to Mendel
Polygenic inheritance – a trait is governed by 2 or more sets of alleles. The dominant
allele effects are additive. Most are environmentally influenced.
Ex. Seed color in wheat, human height, skin color, hair color, eye color Usually
show bell curve distribution
Pleiotropy – one gene affects multiple traits
Ex. SRY gene is associated with male y chromosome. Produces and enzyme and
if enzyme is present you get male genitals, body hair, male fat distribution and
other male traits. Cystic fibrosis is a problem with one gene, but most symptoms
are phenotypic expressions of the problem with the single gene
Multiple alleles in a human population (Ex. Blood types) but individuals only inherit and
express 2 alleles for each gene
Incomplete dominance – the heterozygote has an intermediate phenotype between that of
either homozygote.
Ex. Red and white flowers make pink, curly and straight hair make wavy hair,
Co-dominance – in heterozygote both alleles are fully expressed
Ex. Blood type. Genes for a antigen, b antigen or neither. (Rh determines positive
or negative and it is not codominant)
Epistasis – one gene interferes with the expression of another gene
Ex. Albinos lack melanin. They have melanin genes, but they also have a gene
that interferes with the expression of these alleles.
Environment and the phenotype.
In height nutrition is a major factor
Primroses are white above 32degrees C and red at 24degrees C
Siamese cats and Himalayan rabbits are darker on the ears, nose, paws and tail
where it is cooler. Can replicate cooler regions with ice and the fur will turn dark.
Stress understanding of how Mendel’s Laws are related to meiosis and heredity. Students
should be able to work mono and di hybrid crosses and provide expected ratios. Students
should be able to work these crosses backward to find parent genotypes and phenotypes. I
am more concerned with them getting the correct answer than the method they use to
arrive at it. While there are merits to knowing how to use both punnett squares and laws
of probability I am more concerned with them understanding the logic behind the crosses
and being able to get accurate ratios. They should also know how to work crosses when
the traits are Non-Mendelian and how the ratios change. They also need to understand
how disorders are passed on (autosomal dominant disorders passed via recessive allele
and autosomal recessive disorders on recessive alleles where heterozygotes are carriers).
They don’t need to memorize the genetics of any trait. I will always tell them if the trait
is recessive or co-dominant or polygenic. They should be able to tell me the genetics if I
give them a pedigree though, but they should be able to tell me this not because they
memorized the genetics, but because they understand pedigrees.