Title: Patterns of Inheritance
1Patterns of Inheritance
2Gregor Mendel
- 1822 1884
- Austrian monk
- Experimented with garden peas
- Provided a basis for understanding heredity
3Mendel contd
- 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
4Mendel contd.
- Mendels 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
5Modern Genetics Genes
- 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 organisms genome.
- Controls the physical characteristics of a species
6Modern Genetics - Alleles
- 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
7Modern Genetics Alleles contd
8Modern Genetics Alleles contd
- 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
- Homozygous dominant or heterozygous ? Dominant
Trait - Homozygous recessive ? recessive trait
9Single Gene Inheritance
- Simplest situation
- One gene carries all the information responsible
for one trait - Widows Peak
- Alternative forms of alleles for hairline shape
- Widows peak is dominant to straight hair line
- Wallele for widows peak
- w allele for straight hairline
10Widows peak
11Genotype related to phenotype
12Single Gene Inheritance contd.
- Monohybrid cross
- Looks at inheritance of one trait only
- A punnett squares used to find all possible
combinations of alleles.
13Single Gene Inheritance contd.
- Example 1
- If a homozygous woman with a widows peak
reproduces with a man with a straight hairline,
what kind of hairline will their children have?
14Single Gene Inheritance contd.
- Example
- If two heterozygous parents reproduce what kind
of hairline will their children have?
W w
W WW Ww
w Ww ww
15Single Gene Inheritance contd.
- Genetic Ratios
- Express ratios of possible outcomes
- Genotypic ratio
- Homozygous Dominant Heterozygous Homozygous
Recessive - Phenotypic 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.
16Single Gene Inheritance contd.
- Look back to example 1
- What are the genotypic and phenotypic ratios?
- WW x ww
- Genotypic Ratio
- 0 4 0
- Phenotypic Ratio
- 10
- 100 Widows peak
- Ww x Ww
- Genotypic Ratio
- 1 2 1
- Phenotypic Ratio
- 3 1
- Probability is ¾ Widows Peak
- 75
- Probability ¼ Straight
- 25
17Single Gene Inheritance contd
- Determining Genotype
- No way to distinguish between a homozygous
dominant individual and a heterozygous individual
just by looking - They are phenotypically the same
- Test cross may help us determine
- Used in breeders of plants and animals
- Cross unknown with a recessive individual
- We know one parent genotype, this will help us
determine the other genotype - If there are any offspring produced with the
recessive phenotype, then the dominant parent
must be heterozygous
18Single Gene Inheritance contd
- 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?
19Single Gene Inheritance contd
- 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.
20Homework (WHAT??? Its BIO!)
- Bikini Bottom Beach Genetics
21Independent Assortment
- 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)
22Independent Assortment contd
23Independent Assortment contd
- 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
24Independent Assortment contd
- 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
25Dihybrid Cross
- 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.
26Dihybrid cross contd
- Possible Gametes?
- Genotypic Ratio?
- Phenotypic Ratio?
27Dihybrid Crosses contd
- Determining Ratios
- 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 widows peak ¾
- Probability of short fingers ¾
- What is the probability of widows peak AND short
fingers? - ¾ x ¾ 9/16
28Dihybrid Crosses contd
- Recall from our single trait crosses
- Probability of widows peak ¾ Probability of
short fingers ¾ - Probability of straight hairline ¼ Probability
of long fingers ¼ - Using the product rule
- Probability of widows peak and short fingers
X - Probability of widows peak and long fingers
X 3/16 - Probability of straight hairline and short
fingers ¼ X ¾ 3/16 - Probability of straight hairline and long fingers
¼ X ¼ 1/16
These values are standard for all heterozygous
crosses! You dont need to memorize them, but
should be able to figure them out in your head!
29Dihybrid Crosses contd
- Using the product rule
- Probability of widows peak and short fingers
- Probability of widows peak and long fingers
- Probability of straight hairline and short
fingers - Probability of straight hairline and long fingers
-
30Dihybrid Crosses contd
- 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? x wwss
WS W? ?S ??
wwss WwSs Ws?s ?wSs ?w?s
31Dihybrid Cross contd
- 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?
32Polygenic Inheritance
- 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
33Polygenic inheritance contd
34Polygenic Inheritance contd
- Skin color
- Controlled by many gene pairs and many alleles
- Lets 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
35Polygenic Inheritance contd
- Eye colour is controlled by 3 genes we have
identified - We suspect there are more
- Not a clear dominant and recessive
- Brown allele
36Environmental Influences on Inheritance
- 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
37Coat color in Himalayan rabbits
38Incomplete Dominance
- Incomplete dominance
- Heterozygous individual has a phenotype
intermediate to the two homozygous individuals - Ex Curly-haired Caucasian woman and a
straight-haired Caucasian man produce wavy-haired
children - When 2 wavy-haired people have children, the
phenotypic ratio is 1 curly 2 wavy 1 straight
39Incomplete dominance
40Codominance
- 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
41Codominance contd.
- 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
42Codominance contd.
- 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
43Inheritance of blood type
44Practice Problems
- 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 Mrs. Doe type A Mrs. Jones
type A - Baby 2 type B Mr. Doe type A Mr. Jones
type AB
45Sex-linked inheritance
- Sex chromosomes
- 22 pairs of autosomes, 1 pair of sex chromosomes
- X 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
46Sex-linked inheritance contd.
- X-linked traits
- Red-green colorblindness is X-linked
- 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
47Cross involving an X-linked allele
48Practice Problems
- 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 girls parentage? - What is the genotype of a colorblind male with
long fingers is slong fingers? If all his
children have normal vision and short fingers,
what is the likely genotype of the mother?
49Inheritance of linked genes
- 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
50Inheritance of linked genes
- 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
51Practice Problems
- When AaBb individuals reproduce, the phenotypic
ratio is about 31. 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?