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Chapter 7: Extending
Mendelian Genetics
Notes – part one
Section 1: Chromosomes &
Phenotype
The chromosomes on which genes are located can
affect the expression of traits.
Autosomal Genes & Phenotype
• Autosomal genes are found on autosomes
– Chromosomes 1-22 in humans
– Not on sex chromosomes
• Phenotype is determined by two copies of an
autosomal gene (one from each parent)
– AA, Aa, aa
• Mendel studied autosomal characteristics
• Most traits in humans are determined by autosomal
genes
Autosomal Disorders
• Many genetic disorders are caused by recessive
alleles on autosomes
– GG: healthy
– Gg: healthy but carrier for the disorder
– gg: has disorder
– Example : Tay Sachs disease – autosomal
recessive brain disease in very young children
Autosomal Disorders
• Very few genetic disorders are caused by
dominant alleles on autosomes (some are
however)
– HH and Hh: has disorder
– hh: healthy
– Example – Achondroplastic dwarfism
• DD – death of fetus; Dd – dwarf; dd – normal height
Autosomal Traits – sample problem
• Human traits determined by one autosomal gene
follow the same rules when solving genetics
problems as Mendel’s monohybrid crosses
• Tay Sachs is a recessive and autosomal human
disease. A couple has a baby who has Tay
Sachs disease. Neither parent knows of anybody
in their family who had the disease. Write the
genotypes of the parents and predict the
percentage of their children who will have this
disease using a Punnett square.
• Parents are carriers – genotypes = Tt X Tt
• Punnett square :
•
• Odds of a child inheriting Tay Sachs disease =
• Parents are carriers – genotypes = Tt X Tt
• Punnett square :
•
T
t
T
t
• Odds of a child inheriting Tay Sachs disease =
• Parents are carriers – genotypes = Tt X Tt
• Punnett square :
•
T
t
T
TT Tt
t
Tt tt
• Odds of a child inheriting Tay Sachs disease =
• Parents are carriers – genotypes = Tt X Tt
• Punnett square :
•
T
t
T
TT Tt
t
Tt tt
• Odds of a child inheriting Tay Sachs disease =
¼ or 25%
Sex-Linked Traits
• Genes located on the sex chromosomes
(X or Y) are called sex-linked genes
• Genes on the Y chromosome are responsible for
male characteristics only
• X chromosome has many more genes that affect
many traits
• In males, all sex-linked traits are expressed
– Males are XY
– they only have one copy of each chromosome,
so all alleles are expressed even if they’re
recessive
Some disorders on the X
chromosome
(sex linked disorders)
•
•
•
•
Hemophilia
Muscular Dystrophy
Color blindness
SCID (bubble boy)
How to write genotypes for
sex linked traits
• To represent
that the traits
are on the X
chromosomes,
we write the
dominant or
recessive allele
as a superscript
on the X
Sex Linked Traits – sample problem
• Color blindness is a recessive sex linked
disorder. A color blind man and his wife
have a son who is color blind. The wife is
upset that her husband passed color
blindness on to their son. Use a Punnett
square to show her who she should be
upset with. What percentage of their sons
are predicted to be color blind?
• – genotypes of parents (assume mom is normal)
• Mom – XB XB
• Dad - Xb Y
• Punnett square :
•
• – genotypes of parents (assume mom is normal)
• Mom – XB XB
• Dad - Xb Y
• Punnett square :
XB
•
Xb
Y
XB
• – genotypes of parents (assume mom is normal)
• Mom – XB XB
• Dad - Xb Y
• Punnett square :
•
XB
XB
Xb XBXb XBXb
BY
X
Y
XBY
* A color blind dad
does not pass his
X on to any of his
sons – just the Y
- A color blind son
inherited the color
blindness allele
from his MOM
• – genotypes of parents – mom MUST be a carrier
• Mom – XB Xb
• Dad - Xb Y
• Punnett square :
• – genotypes of parents – mom MUST be a carrier
• Mom – XB Xb
• Dad - Xb Y
• Punnett square :
XB
•
Xb
Y
Xb
• – genotypes of parents – mom MUST be a carrier
• Mom – XB Xb
• Dad - Xb Y
• Punnett square :
•
XB
Xb
Xb XBXb XbXb
BY
X
Y
XbY
* 50% of their sons
will be color blind
50% of their
daughters will be
color blind (girls
can be color blind
but it’s very rare)
X Chromosome Inactivation
• Since females have two X chromosomes, one
of them is randomly turned off in each cell
during embryonic development – leads to
“patchiness” of phenotype for X linked traits
X Chromosome Inactivation
• Because it is random, a female will have
some groups of cells with one X turned off,
and others with the other X turned off
X chromosomes
that are turned
off are visible as
Barr bodies
(one way to
determine if cells
are from a male
or a female)
Section 2: Complex Patterns
of Heredity
Phenotype is affected by many different factors.
• Incomplete dominance - heterozygous genotype
shows a blending of the two homozygous
phenotypes
• Ex. Four o’clock plants
– AA = red flowers
– Aa = pink flowers
– aa = white flowers
• How to solve problems that involve traits that
have alleles that show incomplete dominance
• Pay close attention to genotypes and phenotypes
• Otherwise they are exactly like monohybrid
autosomal crosses
• Sample problem : what percentage of a cross
between two heterozygous four o’clocks are
expected to be pink? Red? White? Color in four
o’clock flowers is determined by alleles that show
incomplete dominance
• Genotypes of parents= Aa X Aa
• Punnett square :
•
• Genotypes of parents= Aa X Aa
• Punnett square :
•
A
A
a
a
• Genotypes of parents= Aa X Aa
• Punnett square :
•
A
a
A
AA Aa
a
Aa aa
• Genotypes of parents= Aa X Aa
• Punnett square :
•
A
a
A
AA Aa
a
Aa aa
• 25% - red AA
• 50% - pink Aa
• 25% - white aa
• Codominance - both alleles are dominant, so
both are fully and separately expressed
• Ex. Human ABO blood type - has multiple
alleles (IA, IB, and i), IA and IB are codominant;
i is recessive to both
genotypes
– IAIA or IAi
– IBIB or IBi
– IAIB
– ii
phenotypes
type A blood
type B blood
type AB blood
type O blood
-Type AB is the universal recipient
-Type O is the universal donor
-A or B antigens must already be present if person is to
receive blood with either antigen in it or results can be
fatal
• How to solve problems that involve codominant
traits
• Pay close attention to genotypes and phenotypes
• Otherwise they are exactly like monohybrid
autosomal crosses
• Sample problem : Two parents with Type A blood
are in a car accident with their baby. At the
hospital they tell the doctors that their baby is
definitely Type A. The doctors give the baby Type
A blood. The baby quickly gets very sick and
needs to go to the ICU. What happened? Explain
using a Punnett square.
• Genotypes of parents= IAi X IAi
• Punnett square :
•
• Genotypes of parents= IAi X IAi
• Punnett square :
•
IA
IA
i
i
• Genotypes of parents= IAi X IAi
• Punnett square :
•
IA
i
IA
IAIA IAi
i
IAi
ii
• Genotypes of parents= IAi X IAi
• Punnett square :
•
IA
i
IA
IAIA IAi
i
IAi
ii
• 25% - IAIA
• 50% - IAi
• 25% - ii
Type A
Type A
Type O
-The baby is Type O –
each parent is a
heterozygote
- this is why doctors
always test before
starting any blood
transfusion.
another example of codominance : Sickle Cell
Anemia
symptoms : weakness, dizzy spells, blood flow issues
genetic cause : codominant allele
genotypes/phenotypes :
HAHA= normal RBC’s
HAHS= some sickle RBC’s, not enough to make
person sick; resistance to malaria
HSHS= sickle cell sufferer
Distribution of Sickle Cell Anemia mirrors that
of Malaria - people who are heterozygous for
Sickle Cell Anemia are more resistant to
malaria (heterozygote advantage)
• Polygenic traits - traits produced by two or
more genes, leading to a range of phenotypes
• Ex. human height, human eye color
In a polygenic trait, one gene may be epistatic,
meaning it can interfere with the expression of
the other genes
– Ex. albinism in mammals
Genes are often affected by
the Environment
• Phenotype is usually a mixture of genes & the
environment
• Ex. genes determine your potential height but
nutrition and diet may limit your actual height
The gender of sea turtles depends
on environmental temperatures
when they are developing
-Females are more prevalent when
temperatures are warm
-More males are produced in cool
temperatures