Transcript Dihybrid crosses and gene linkage

HL Genetics Topic 10.2
Assessment Statements
 10.2.1 Calculate and predict the genotypic and phenotypic





ratio of offspring of dihybrid crosses involving unlinked
autosomal genes.
10.2.2 Distinguish between autosomes and sex
chromosomes.
10.2.3 Explain how crossing over between non-sister
chromatids of a homologous pair in prophase I can result
in an exchange of alleles.
10.2.4 Define linkage group.
10.2.5 Explain an example of a cross between two linked
genes.
10.2.6 Identify which of the offspring are recombinants in a
dihybrid cross involving linked genes.
Dihybrid crosses
 Mendel’s Peas
 Seed shape – some round, others wrinkled (allele for
round is dominant)
 Seed colour – some green, others yellow (allele for
yellow is dominant)
 Mendel crossed true breeding plants with each other
 One parent: homozygous dominant for both traits
(round and yellow seeds) RRYY
 Other parent: homozygous recessive for both traits
(wrinkled and green) rryy
When both parents are homozgous – all the F1 offspring
are the same genotype and phenotype
 R = allele for round peas
Parent
Round
phenotypes yellow
Green
wrinkled
 r = allele for wrinkled peas
Parent
genotypes
RRYY
rryy
Parent
gametes
RY
ry
F1
genotypes
RrYy
 Y = allele for yellow peas
 y = allele for green peas
F1
Round
phenotypes yellow
Cross the F1 double heterozygous
 Allowing heterozygous offspring to self-pollinate
Phenotype
Ratio is:
9
3
3
1
Dihybrid Phenotype Ratios
Homozygous parents
AABB
x aabb
All F1 offspring the same
Heterozygous parents
AaBb x AaBb
9:3:3:1 phenotype ratio
9AB 3Ab 3aB 1ab
A new shuffling of the alleles has created a new combination which
does not match either of the parents’ genotypes. Recombinants
Autosomes and sex chromosomes
 Sex chromosomes: X and Y (one pair)
 Autosomes: any chromosome not X or Y (22 pairs)
 Sex-linked gene





is located on a sex chromosome.
Autosomal gene is located on one of the autosomes.
On which type of chromosome is the gene for
haemophilia found?
So, the gene is known as ___.
On which type of chromosome is the gene for protein
production in the pancreas found?
So, the gene is known as ___.
Exchange of alleles by crossing over
 Two non-sister chromatids can swap segments of their
DNA during prophase I of meiosis.
 This increases genetic variety of chromosomes in
gametes
There are now thought to be 20,500 human
genes on 23 chromosomes
Source: Broad Institute of MIT and Harvard (2008, January 15). Human Gene Count Tumbles
 There must be approx 1000 genes on each chromosome.
Linkage group
 Any two genes which are found on the same
chromosome are said to be linked to each other.
 Linked genes are usually passed on to the next
generation together.
 Linkage group - groups of genes on the same
chromosome inherited together
 Linked genes are the exception to Mendel’s law of
independent assortment
Linked genes
Fruit fly gene for body
color is in the same
linkage group as the gene
for wing length
Notation of linked genes is
G L
G L
G L
G L
Alleles are
G – grey body
g – black body
L – long wings
l – short wings
The two horizontal bars
symbolize homologous
chromosomes and that
the locus of G is on the
same chromosome as L
Offspring of a dihybrid cross
A cross between homozygous fruit flies
GGLL x ggll
Grey body & long wings x Black body, short wings
F1 flies will be all heterozygous for both of the traits
GgLl
Phenotype : Grey body & long wings
If these F1 heterozygotes were allowed to reproduce together
the ratio of phenotypes produced would be expected to be
9:3:3:1
Recombinants
After the F2 flies were identified the phenotypes were;
1600
Grey
long
wing
Grey
short
wing
Black
long
wing
Black
short
wing
1020
210
200
170
300
100
This is far from the expected 9:3:3:1 ratio
which should give
900
300
This is a sign that the genes are on the same chromosome (genes are linked)
Offspring of a dihybrid cross
using linkage notation
A cross between homozygous fruit flies
GL
gl
GL x gl
Grey body & long wings x Black body, short wings
F1 flies will be all heterozygous for both of the traits
GL
gl
Phenotype : Grey body & long wings
If these F1 heterozygotes were allowed to reproduce together the ratio of
phenotypes produced would be expected to be
9:3:3:1
Offspring of a dihybrid cross
using linkage notation
G L
G L
If these F1 heterozygotes make gametes
The gametes will be
GL or gl
unless crossing over happens – because the genes are on
the same chromosome
The F1 will contain more than expected of these
genotypes
GL
GL
gl
gl
GL
gl
GL
gl
Offspring of a dihybrid cross
using linkage notation
G L
G L
If The F1 will contain more than expected of these
genotypes
1600
Grey
long
wing
Grey
short
wing
Black
long
wing
Black
short
wing
1020
210
200
170
GL
GL
gl
gl
GL
gl
GL
gl
The Grey short wing and black long wing phenotypes
Are Recombinants -
Topic 10.3
Assessment Statements
 10.3.1 Define polygenic inheritance.
 10.3.2 Explain that polygenic inheritance can
contribute to continuous variation using two
examples, one of which must be human skin colour.
Polygenic Inheritance
when two or more genes influence
the expression of one trait
Eg Skin Colour, Height
Continuous and discontinuous variation
 When an array of possible phenotypes can be
produced, it is called continuous variation
 Examples: skin color, height, body shape, and
intelligence
 These traits are also influenced by environmental
conditions
 When only a number of phenotypes can be produced,
it is called discontinuous variation
 Examples: earlobe attachment, blood group
Graphical representation
Continuous variation
Discontinuous
Height in humans
60
50
40
30
Frequency
20
10
0
4
6
variation
Blood Type
45
40
35
30
25
20
15
10
5
0
% of
populatio
n
A
B
AB
O
Eye Color
 Iris is made up of zones,
rings, streaks or speckles of
different colored pigments
with varying intensities
 What color are your eyes,
really?
 Since there is so much
variety, eye color must be
influenced by multiple
alleles and has continuous
variation.
Skin color
How does the existence of multiple alleles controlling
skin colour result in the appearance of many different
shades of skin colour in humans?
Thoughts
 How do people of varying degrees of skin color
relocated to parts of the world that receive differing
amounts of sunlight get vitamin D? How do others
fight off the sun?
 Should there be equal esteem for all humans?
 Why is human diversity so often used to divide and
discriminate, rather than be appreciated, respected,
and celebrated?
Polygenic inheritance of color in
wheat.
Kernal color in wheat is
determined by two genes.
A range of colors occur,
from white to dark red,
depending on the
combinations of alleles.
Dark red plants are
homozygous AABB and
white plants are
homozygous aabb.
Crossing individuals with the phenotype extremes
yield offspring that are a 'blend' of the two parents.
When these homozygous phenotypes are crossed
AABB x aabb
Dark x white
the F1 offspring are all double heterozygous
AaBb.
But what happens when the two double
heterozygous genotypes are crossed?
 Parent Phenotypes: all brown
 Genotypes: AaBb x AbBb
 Punnet square:
AB
Ab
aB
ab
AB
AABB AABb
AaBB
AaBb
Ab
AABb
AAbb
AaBb
Aabb
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
There is no blending in
the offspring. Offspring
can be more extreme
than either parent
There are grades of
colour – evidence of
polygenic inheritance.
Skin color
How does the existence of multiple alleles controlling
skin colour result in the appearance of many different
shades of skin colour in humans?
Human skin colour is controlled by
multiple alleles (and the environment)
 It is known that at least three genes control skin color,
let’s call them genes A, B, and C.
 Someone who is AABBCC would have very dark skin
color and someone who is aabbcc would have very
light skin color.
 If they got married and had children, their children
would all be AaBbCc and have mid-brown skin.
 If two of those people would get married and have
children, the Punnett square would look like the one
above.