Transcript Mendelian Genetics - hills

23-1
PATTERNS OF INHERITANCE
Chapter 23
Gregor Mendel
23-2




1822 – 1884
Austrian monk
Experimented with
garden peas
Provided a basis for
understanding
heredity
Mendel
cont’d
23-3

Published a paper in 1866 stating that parents
pass discrete heritable factors on to their offspring
 Factors
retain individuality generation after generation
 Identified that each trait is inherited by a pair of
factors, one from each parent
 One
form of a factor may be dominant over an alternative
form
 Reasoned that each egg and sperm must contain only 1
copy of a factor for each trait
Mendel cont’d.
23-4

Mendel’s law of segregation
Each individual has two factors (genes) for each trait
 The factors segregate (separate) during the formation of
gametes
 Each gamete contains only one factor from each pair of
factors
 Fertilization gives each new individual 2 factors for each
trait

Modern Genetics – Genes
23-5



Sections of chromosomes which give instructions for
one characteristic or protein.
Located at the same point or locus, on each member
of a homologous pair
All together make up the organism’s genome.
 Controls
the physical characteristics of a species
Modern Genetics - Alleles
23-6

Alternative forms of the same gene on each chromosome


One allele comes from each parent
Dominant allele




Masks other traits present
Only 1 dominant allele needs to be present for a certain trait to be
expressed
Represented by a capital letter
Recessive Allele


2 copies of the recessive allele need to be present for trait to be
expressed
Represented by a lower case letter
Modern Genetics – Alleles cont’d
23-7

Gene locus

Fig. 23.2
Modern Genetics – Alleles cont’d
23-8

Genotype
 Genetic
composition of a specific trait
 Homozygous
dominant – 2 dominant alleles
 Heterozygous – 1 dominant allele and 1 recessive allele
 Homozygous recessive 2 recessive allele

Phenotype
 Physical
expression of a specific trait
dominant or heterozygous  Dominant Trait
 Homozygous recessive  recessive trait
 Homozygous
Single Gene Inheritance
23-9

Simplest situation
 One
gene carries all the information responsible for
one trait

Widow’s Peak
 Alternative
forms of alleles for hairline shape
 Widow’s peak is dominant to straight hair line
 W=allele
for widow’s peak
 w= allele for straight hairline
Widow’s peak
23-10

Fig. 23.3
Genotype related to phenotype
23-11

Table 23.1
Single Gene Inheritance cont’d.
23-12

Monohybrid cross
 Looks
at inheritance of one trait only
 A punnett squares used to find all possible combinations
of alleles.
Single Gene Inheritance cont’d.
23-13

Example 1:
 If
a homozygous woman with a widow’s peak
reproduces with a man with a straight hairline, what
kind of hairline will their children have?
Single Gene Inheritance cont’d.
23-14

Example:
 If
two heterozygous parents reproduce what kind of
hairline will their children have?
W
w
W WW Ww
w Ww ww
Single Gene Inheritance cont’d.
23-15

Genetic Ratios
 Express
ratio’s of possible outcomes
 Genotypic

Homozygous Dominant: Heterozygous: Homozygous Recessive
 Phenotypic



ratio
ratio
Dominant trait : recessive trait
Often expressed as probability
The probability resets and is the same for each pregnancy!
 Having one child with a trait has no effect on future children.
Single Gene Inheritance cont’d.
• Look back to example
1
– What are the
genotypic and
phenotypic ratio’s?
– WW x ww
• Genotypic Ratio
– 0:4:0
• Phenotypic Ratio
– Ww x Ww
• Genotypic Ratio
– 1: 2: 1
• Phenotypic Ratio
– 3:1
– Probability is ¾
Widow’s Peak
» 75%
– Probability ¼ Straight
» 25%
– 1:0
– 100% Widow’s peak
23-16
Single Gene Inheritance cont’d
23-17

Determining Genotype
 No
way to distinguish between a homozygous dominant
individual and a heterozygous individual just by looking
 They
 Test
are phenotypically the same
cross may help us determine
 Used
in breeders of plants and animals
 Cross unknown with a recessive individual

 If
We know one parent genotype, this will help us determine the
other genotype
there are any offspring produced with the recessive
phenotype, then the dominant parent must be heterozygous
Single Gene Inheritance cont’d
23-18

Example :
In rats, large ears is dominant to small ears. A rat
breeder has a female rat with large ears, which she
breeds with a male rat with small ears. In the first
litter, all rats are born with large ears.
 What

is the genotype of the female rat?
In a second litter from the same parents, 4 baby
rats have large ears, one has small ears.
 What
is the genotype of the female rat?
Single Gene Inheritance cont’d
23-19

Practice problems

Both a man and a woman are heterozygous for freckles. Freckles
are dominant over no freckles. What is the chance that their child
will have freckles?

Both you and your sibling have attached ear lobes, but your
parents have unattached lobes. Unattached earlobes (E) are
dominant over attached (e). What are the genotypes of your
parents?

A father has dimples, the mother of his children does not, and all
5 of their children have dimples. Dimples (D) are dominant over
no dimples (d). Give the probable genotypes of all persons
concerned.
Homework (WHAT??? It’s BIO!)
23-20

Bikini Bottom Beach Genetics
Independent Assortment
23-21

Mendel reasoned from the results of his pea plant
crosses that each pair of factors assorts
independently into gametes
 Each
trait is passed down individually. The allele you
receive for any one gene is not related to any other
alleles you receive.
 We
can now explain this through independent alignment
and crossing over in meiosis. (Fig 23.6)
Independent Assortment
23-22

Fig. 23.6
cont’d
Independent Assortment
cont’d
23-23

Called the Law of Independent Assortment
 Each
pairs of factors assorts independently (without regard
to how the others separate)
 All possible combinations of factors can occur in the gametes
Independent Assortment
cont’d
23-24

Practice problems

For each of the following genotypes, give all possible gametes






WW
WWSs
Tt
Ttgg
AaBb
For each of the following, state whether the genotype or a
gamete is represented




D
Ll
Pw
LlGg
Dihybrid Cross
23-25


Punnet squares, considering two-trait crosses at one
time.
Example:

The traits for hairline and finger length are both single gene
traits. As before, widows peak is dominant over straight
hairline. Having short fingers is considered dominant over
long fingers.

Two parents who are both heterozygous for both traits have
children.

Determine the genotypic and phenotypic ratios for their children.
Dihybrid cross
23-26

Possible Gametes?

Genotypic Ratio?

Phenotypic Ratio?
cont’d
Dihybrid Crosses cont’d
23-27

Determining Ratio’s
 Product
rule of probability
 The
chance of 2 or more independent events occurring
together is the product of their chance of occurring
separately
 In
our example:
 Probability
of widow’s peak = ¾
 Probability of short fingers= ¾
 What is the probability of widow’s peak AND short fingers?

¾ x ¾ = 9/16
Dihybrid Crosses cont’d
23-28

Recall from our single trait crosses



Probability of
Probability of
Probability of
Probability of
widow’s peak = ¾
short fingers= ¾
straight hairline= ¼
long fingers= ¼
These values are standard for
all heterozygous crosses! You
don’t need to memorize them,
but should be able to figure
them out in your head!
Using the product rule

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Probability of
Probability of
Probability of
3/16
Probability of
1/16
widow’s peak and short fingers = X =
widow’s peak and long fingers = X = 3/16
straight hairline and short fingers = ¼ X ¾ =
straight hairline and long fingers = ¼ X ¼ =
Dihybrid Crosses cont’d
23-29

Using the product rule

Probability of widow’s peak and short fingers

Probability of widow’s peak and long fingers

Probability of straight hairline and short fingers

Probability of straight hairline and long fingers
Dihybrid Crosses cont’d
23-30

Two-trait test cross
 Cross
an individual with the dominant phenotype for
each trait with an individual with the recessive
phenotype of both traits
 W?S?
wwss
x wwss
WS
WwSs
W?
Ws?s
?S
?wSs
??
?w?s
Dihybrid Cross cont’d
23-31

Attached earlobes are recessive, What genotype do
children have if one parent is homozygous for earlobes and
homozygous dominant for hairline, and the other is
homozygous dominant for unattached earlobes and
homozygous recessive for hairline?

If an individual from this cross reproduces with another of
the same genotype, what are the chances that they will have
a child with a straight hairline and attached earlobes?

A child who does not have dimples or freckles is born to a
man who has dimples and freckles (both dominant traits)
and a woman who does not. What are the genotypes of all
persons concerned?
Polygenic Inheritance
23-32


Not fully understood by geneticists.
Generally:
One trait controlled by 2 or more genes at different loci
 The higher the number of dominant alleles you possess, the
stronger the expression of the trait.


Result is a continuous range of phenotypes
Distribution resembles a bell curve
 The more gene pairs involved, the more continuous the
pattern of variation
 Ex: human height, skin pigmentation, eye colour

Polygenic inheritance cont’d
23-33

Fig. 23.9
Polygenic Inheritance cont’d
23-34

Skin color
 Controlled
by many
gene pairs and many
alleles
 Let’s assuming a simple
model of two alleles at
2 loci
A
and B
 If two heterozygous
parents have children,
children can range from
very light to very dark
Genotype
Phenotype
AABB
Very Dark Skin
AABb or AaBB
Dark Skin
AaBb, AAbb, or
aaBB
Medium brown skin
Aabb, or aaBb
Light Skin
aabb
Very light skin
Polygenic Inheritance cont’d
23-35

Eye colour is controlled
by 3 genes we have
identified
 We
suspect there are
more


Not a clear dominant
and recessive
Brown allele
Environmental Influences on Inheritance
23-36

Environment can influence gene expression and
therefore phenotype
 Ex:
sunlight exposure on skin; coat color in Himalayan
rabbits

Human twin studies
 Polygenic
traits are most influenced
 “nature vs. nurture”
 Identical twins separated at birth are studied
 If
they share a trait in common even though raised in
different environments, it is likely genetic
Coat color in Himalayan rabbits
23-37

Fig. 23.10
Incomplete Dominance
23-38

Incomplete dominance
 Heterozygous
individual has a phenotype intermediate
to the two homozygous individuals
 Ex: Curly-haired Caucasian woman and a straighthaired Caucasian man produce wavy-haired children
 When
2 wavy-haired people have children, the phenotypic
ratio is 1 curly: 2 wavy: 1 straight
Incomplete dominance
23-39

Fig. 23.11
Codominance
23-40

Multiple allele inheritance


The gene exists in several allelic forms, but each person still has
only 2 of the possible alleles
Occurs when both alleles are equally expressed
 Ex:
type AB blood has both A antigens and B antigens
on red blood cells
Codominance cont’d.
23-41

ABO blood types
 IA
= A antigens on RBCs
 IB = B antigens on RBCs
 i = has neither A nor B
antigens on RBCs
 Both IA and IB are
dominant over I, IA and
IB are codominant
Phenotype
Genotype
A
IAIA or IAi
B
IBIB or Ibi
AB
IAIB
O
ii
Codominance cont’d.
23-42

Paternity testing- ABO blood groups often used
 Can

disprove paternity but not prove it
Rh factor- another antigen on RBCs
 Rh
positive people have the antigen
 Rh negative people lack it
 There
are multiple alleles for Rh negative, but all are
recessive to Rh positive
Inheritance of blood type
23-43

Fig. 23.12
Practice Problems
23-44
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
A polygenic trait is controlled by three pairs of alleles. What are
the two extreme genotypes for this trait?
What is the genotype of the lightest child that could result from a
mating between two medium-brown individuals?
A child with type O blood is born to a mother with type A blood.
What is the genotype of the child? The mother? ‘what are the
possible genotypes of the father?
From the following blood types determine which baby belongs to
which parents:


Baby 1 type O
Baby 2 type B
Mrs. Doe type A
Mr. Doe type A
Mrs. Jones type A
Mr. Jones type AB
Sex-linked inheritance
23-45

Sex chromosomes
 22
X
pairs of autosomes, 1 pair of sex chromosomes
and y


In females, the sex chromosomes are XX
In males, the sex chromosomes are XY
 Note that in males the sex chromosomes are not homologous
 Traits
controlled by genes in the sex chromosomes are
called sex-linked traits
X
chromosome has many genes, the Y chromosome does not
Sex-linked inheritance cont’d.
23-46

X-linked traits
 Red-green
colorblindness is Xlinked
 The
X chromosome has
genes for normal color
vision


XB = normal vision
Xb – colorblindness
Genotypes
Phenotypes
XBXB
female with normal vision
XBXb
carrier female, normal vision
XbXb
colorblind female
XBY
male with normal vision
XbY
colorblind male
Cross involving an X-linked allele
23-47

Fig. 23.13
Practice Problems
23-48




Both the mother and the father of a colorblind male appear to be normal.
From whom did the son inherit the allele for colorblindness? What are the
genotypes of the mother, father, and the son?
A woman is colorblind. What are the chances that her son will be
colorblind? If she is married to a man with normal vision, what are the
chances that her daughters will be colorblind? Will be carriers?
Both the husband and the wife have normal vision. The wife gives birth to a
colorblind daughter. Is it more likely the father had normal vision or was
colorblind? What does this lead you to deduce about the girl’s parentage?
What is the genotype of a colorblind male with long fingers is s=long
fingers? If all his children have normal vision and short fingers, what is the
likely genotype of the mother?
Inheritance of linked genes
23-49


The sequence of individual genes on a chromosome is
fixed because each allele has a specific locus
All genes on a single chromosome form a linkage group
When linkage is complete, a dihybrid produces only 2 types
of gametes
 Any time traits are inherited together, a linkage group is
suspected
 If very few recombined phenotypes appear in offspring,
linkage is also suspected

Inheritance of linked genes
23-50


Crossing over between
2 alleles of interest,
can result in 4 types of
gametes
Occurrence of crossing
over can indicate the
sequence of genes on
a chromosome
 More
frequent
between distant genes
Fig. 23.14
Practice Problems
23-51


When AaBb individuals reproduce, the phenotypic ratio
is about 3:1. What ratio was expected? What may
have caused the observed ratio?
The genes for ABO blood type and for fingernails are
on the same homologous pair of chromosomes. In an
actual family, 45% of offspring have type B blood and
no fingernails, and 45% have type O blood and
fingernails; 5% have type B blood and fingernails, and
5% have type O blood and no fingernails. What
process accounts for the recombinant phenotypes?