Download GgNn - Blue Valley Schools

Other
Patterns of
Inheritance
Other Patterns of Inheritance are
Exceptions to Mendel’s Laws
Law of Dominance: if the two alleles at a locus
differ, then one, the dominant allele, determines the
organism′s appearance; the other, the recessive
allele, has no noticeable effect on the organism′s
appearance
Law of Segregation: the two alleles for a heritable
character separate (segregate) during gamete
formation and end up in different gametes
Law of Independent Assortment: each pair of alleles
segregates independently of other pairs of alleles
during gamete formation
Codominance
• The heterozygote expresses both
alleles at the same time.
• In this example, the heterozygous
cattle are roan. They express
both hair colors at the same time.
Red
“Roan”
White
R
R
CC
x
R
R
CC
W
W
C C
x
W
W
C C
R
W
CC x
R
W
CC
R
R
CC
x
R
W
CC
R
R
CC x
W
W
C C
R
W
CC
x
W
W
C C
Multiple Alleles
• For many genes, there are
several alleles exist in the
population.
• This expands the number of
possible genotypes and
phenotypes.
ABO Blood Types
A
B
I I
x ii
x
Dihybrid Cross
• The mating of two organisms that
differ in two characters.
Sample Two-Factor
Cross Problem
• In pea plants, green pods (G) are
dominant over yellow pods (g), and
smooth pods (N) are dominant over
constricted pods (n). A plant
heterozygous for both traits is
crossed with a plant that has yellow
constricted pods. What are the
probable genotypes and phenotypes
for this cross?
Step 1
• Choose letters to represent the
genes in the cross.
G = green pods
g = yellow pods
N = smooth pods
n = constricted pods
Step 2
• Write the genotypes of the
parents
GgNn X ggnn
Step 3
• Determine the possible gametes
that the parents can produce
(FOIL)
GgNn
GN Gn gN gn
ggnn
gn gn gn gn
Step 4
• Enter the possible gametes at the
top and side of the Punnett
square.
GgNn X
ggnn
gn
gn
gn
gn
GN
Gn
gN
gn
Step 5
• Complete the Punnett square by
writing the alleles from the
gametes in the appropriate boxes.
GgNn X
ggnn
GN
Gn
gN
gn
gn
GgNn
Ggnn
ggNn
ggnn
gn
GgNn
Ggnn
ggNn
ggnn
gn
GgNn
Ggnn
ggNn
ggnn
gn
GgNn
Ggnn
ggNn
ggnn
Step 6
• Determine the genotypes and
phenotypes of the offspring.
Genotypes
•
•
•
•
GgNn 4/16 – 25%
Ggnn 4/16 – 25%
ggNn 4/16 – 25%
ggnn 4/16 – 25%
Phenotypes
•
•
•
•
Green smooth:
4/16 – 25%
Green constricted: 4/16 – 25%
Yellow smooth:
4/16 – 25%
Yellow constricted: 4/16 – 25%
Incomplete Dominance
• The heterozygote has a
phenotype that is intermediate
between the phenotypes of the
two homozygotes.
• Example: Petal color in certain
flowers.
Epistasis
•
•
Effects of one gene override or
mask the phenotype of a second
gene.
Epistasis is not dominance.
Compare the definitions:
Epistasis: One gene masks the expression of a
different gene for a different trait
Dominance: One allele masks the expression of
another allele of the same gene
Example of Epistasis
• Labrador retrievers can be black,
brown, or yellow. Two genes control
this.
• One gene influences melanin production
o B (black) is dominant to b (brown)
• One gene influences melanin deposition
o E (full deposition) is dominant to
(reduced deposition)
e
Sex-Linked Inheritance
• Genes located on sex chromosomes
produce different patterns in males
and females.
• Females generally have two alleles for
these genes.
• Males generally have only one allele.
• If a male inherits a sex-linked recessive
allele from his mother, the allele will
be expressed.
Red/Green Color Blindness
X Chromosome Inactivation
• At a certain point in the embryonic Barr body
development of every female
mammal, one of the two X
chromosomes in each cell inactivates
by supercoiling into a structure
known as a Barr Body.
• This irreversible process leaves only
one active X chromosome in each
cell, and which X chromosome
undergoes inactivation is random
with respect to the cell lineages that
result from future cell divisions.
• If the female is heterozygous, an
interesting pattern can result.
In cats, one of several
genes controlling fur color
is located on the X
chromosome. The gene
has two alleles. One form
of the gene codes for
orange fur (XB), and the
other form codes for black
fur (Xb). The orange allele
is dominant to the black
allele. Ordinarily, this
would mean that an
animal inheriting one copy
of each gene should have
orange fur. However, a
heterozygous female cat
(XBXb) will not be orange.
Instead, her coat will be a
patchwork of orange and
black, a condition known
as tortoiseshell. This
pattern is due to the
random nature of X
chromosome inactivation.
World’s first cloned cat!
Rainbow
CC
CC and her surrogate mother
Pleiotropy
• The control by a single gene of
several distinct and seemingly
unrelated phenotypic effects.
o This occurs because the protein products of
most genes are expressed in more than one
cell type, meaning they participate in more
than one physiological or developmental
process.
• Example: sickle cell anemia
Sickle Cell Anemia
Develops in persons carrying two
defective alleles for a blood
protein, beta-hemoglobin.
Mutant beta-hemoglobins are
misaligned inside a blood cell and
cause misshapen red blood cells
at low oxygen concentrations.
Deformed blood cells impair
circulation. Impaired circulation
damages kidneys and bone. In
this case, the gene defect itself
only affects one tissue, the
blood. The consequences of that
defect are found in other tissues
and organs.
Polygenic
Inheritance
• Multiple genes
influence a
single trait.
Continuous variation
• Most traits show a range of
variation rather than
distinct either/or types
• This occurs when multiple
genes and environmental
factors influence the trait’s
expression
• Continuous variation is
often described with
frequency distribution
tables.
Example using actual data
Autosomal Linkage
• Genes located on the same
chromosome that tend to be
inherited together in genetic
crosses are said to be linked
genes.
• When genes are linked, they do
not assort independently.
Why did Mendel miss it?
• Mendel did not notice linkage
because seven of the
characters he studied in the
garden pea were located on
different (non-homologous)
chromosomes.
• Seed color and flower color
are in the same chromosome,
but they lie so far apart that
they almost always get
separated by crossing over
and assort independently.
Calculating Map Distance
• When genes are linked, the offspring of a testcross
involving an individual heterozygous for each of two
genes will not exhibit a 1:1:1:1 phenotypic ratio
expected for independently assorting genes. Instead,
there will be an excess of the parental phenotypes.
• Results of such testcrosses can be used to calculate the
map distance between the two genes involved.
• Map distance is calculated from the formula for
recombination frequency:
recombination frequency = number of recombinants x 100
(map units)
total offspring
Sample Problem
Sample Problem
Gene Mapping Problem
•
•
•
•
•
•
•
Genes A-B 10 map units
Genes B-C 3 map units
Genes D-B 5 map units
Genes C-E 2 map units
Genes A-D 5 map units
Genes B-E 5 map units
Genes D-E 10 map units
Gene Mapping Problem