Final Concepts for Chapter 11
Mendelian Genetics
Allele
Dominant
Recessive
P-generation
F1
generation
F2 generation
Law of independent
assortment
Law of segregation
Chromosomes
Pure breed
Trait
•Codominance
•Complete dominance
•Dihybrid cross
•Genotype
•Genotypic ratio
•Heterozygous
•Homozygous
•Incomplete dominance
•Monohybrid cross
•Phenotype
•Phenotypic ratio
•Probability
•Punnett square
•Testcross
•Expected/predicted
results
•Actual/observed
results
•Karyotype
•Amniocentesis
•Linked genes
•Sex-linked disorders
•Autosomal disorders
Independent Assortment vs. Linked
Genes
Mendel did not know about chromosomes
when he proposed the Law of
Independent Assortment.
 The pea traits he studied happened to be
located on different chromosomes – so
they did assort independently.

Independent Assortment vs. Linked
Genes

Question: How many traits do you have?

Question: How many chromosomes (per
cell) do you have?

Question: Is it possible to have only one
trait per chromosome?
– No, lots of genes are carried or linked
together on the same chromosome.
Independent Assortment vs. Linked
Genes

Do the punnett square for the following
cross – assume independent assortment.
Cross two heterozygous tall, heterozygous
red flowered plants
T=tall
R=red flower
t= short
r = white flower
Independent Assortment vs. Linked
Genes
What is the
phenotypic ratio?
TR
TR
Tr
tR
tr
TtRr x TtRr
Tr
tR
tr
TTRR
TTRr
TtRR
TtRr
TTRr
TTrr
TtRr
Ttrr
TtRR
TtRr
ttRR
ttRr
TtRr
Ttrr
ttRr
ttrr
Independent Assortment vs. Linked
Genes
9:3:3:1 ratio
9
3
3
1
=
=
=
=
tall and red
tall and white
short and red
short and white
PROBABILITY:
From this cross, 48 offspring
were produced.
1. How many offspring would
you expect to be tall and
red?
2. How many would expect
to be tall and white?
3. How many would you
expect to be short and
white?
Independent Assortment vs. Linked
Genes
Now, do the following cross BUT the genes
for tallness and red flowers are linked.
Cross two heterozygous tall,
heterozygous red flowered plants
T=tall
t= short
R=red flower
r = white flower
Independent Assortment vs. Linked
Genes

Hint
TtRr X TtRr
T
t
R
r
Is it possible to
produce a Tr
gamete?
Independent Assortment vs. Linked
Genes
TtRr X TtRr
TR
TR
tr
TTRR
TtRr
tr
TtRr
ttrr
What is the
phenotypic ratio?
3:1
3 = Tall and Red
1 = Short and
white
Independent Assortment vs. Linked
Genes

So… out of the 48 offspring, if the genes
are linked, how many would be
– 1. tall and red?
– 2. tall and white?
– 3. short and red?
– 4. short and white?
Answer:
tall and red = 36
short and white = 12
EXPECTED
RESULTS!
tall/white = 0
short/red = 0
Independent Assortment vs. Linked
Genes
Is it possible for our Actual Results to
show any flowers that are tall/white or
short/red?
Yes – how?
Crossing over
Crossing over occurs in meiosis
Pieces of the chromosomes actual switch
places.
Complete vs Incomplete
Dominance
Codominance – the alleles are
equally dominant
Roan Cow
Human Blood Type
Sex-linked Traits

Traits carried on the X chromosome
Fill in the genotypes
on the pedigree.
Autosomal disorders
Disorders carried on non-sex
chromosomes (first 22 pairs)
 Some are autosomal dominant

– Huntington’s disease

Most are autosomal recessive
– Sickle-cell anemia
– Cystic fibrosis
Question: How do you know if the pure
bred dog you just paid big bucks for is
actually pure?
GG?
Gg?
Test Cross
Cross using a homozygous recessive
individual with a dominant individual to
determine if the dominant individual is
heterozygous or homozygous dominant
(pure)
 Why use a homozygous recessive
individual?

Test Cross
Do the punnett squares for each case:
GG x gg
Gg x gg
Test Cross

All offspring produced should show the
dominant characteristics if the dominant
parent is pure (GG) for the trait.
Class Quiz – all or nothing!
Key: T = tall and t = short
1. Cross a heterozygous tall plant with a
homozygous dominant tall plant. Show
punnett square ONLY.
2. Cross a heterozygous tall plant with a
short plant. Show punnett square and
PHENOTYPE ratio.

A man and woman marry.
They have five children, 2 girls
and 3 boys. The mother is a
carrier of hemophilia, an Xlinked disorder. She passes
the gene on to two of the boys
who died in childhood and one
of the daughters is also a
carrier. Both daughters marry
men without hemophilia and
have 3 children (2 boys and a
girl). The carrier daughter has
one son with hemophilia. One
of the non-carrier daughter’s
sons marries a woman who is
a carrier and they have twin
daughters. What is the
percent chance that each
daughter will also be a
carrier?




Hemophilia is a
sex-linked trait.
Draw the
pedigree.
Do a punnett
square cross for
each of the
couples to
determine….
What is the
genotype ratio?
What is the
phenotype
ratio?
Answers to front page of quiz
1 point each
Part A
1.M
2. H
3. K
4. A
5. B
6. C
7. E
8. F
9. D
10. I

Answers to front page of quiz
1 point each
Part B
1. d
2. b
3. a
4. h
5. m
6. e
7. c
8. k
9. i
10. f

A
G
g
Answers to the back
3pts
G
g
GG
Gg
Gg
gg

Gg X Gg

Phenotype Ratio:
– 3:1
– 3= green eyes
– 1 = yellow eyes
B
Answers to the back
3pts
Aa X Aa
A
a
What is the phenotypic ratio?
A
a
AA
Aa
Aa
aa
3:1
3 = Big Ears
1 = Small Ears
Or for the G’s
3:1
3= green eyes
1 = yellow eyes
A
Answers to the back
2 points
Bb X bb
b
b
B
b
Bb
bb
Bb
bb
B
Answers to the back
2 points
Tt X tt
T
t
t
t
Tt
tt
Tt
tt
A
1 pt. for the correct genotypes to set up the cross
½ pt. for each correct gamete combo
=6 pts. Total
bbgg x BbGg
bg
BG
Bg
bG
bg
bg
bg
bg
BbGg
BbGg
BbGg
BbGg
Bbgg
Bbgg
Bbgg
Bbgg
bbGg
bbGg
bbGg
bbGg
bbgg
bbgg
bbgg
bbgg
B
1 pt. for the correct genotypes to set up the cross
½ pt. for each correct gamete combo
=6 pts. Total
Bbgg x BbGg
BG
Bg
Bg
bg
bg
Bg
bG
bg
BBGg
BBgg
BbGg
Bbgg
BBGg
BBgg
BbGg
Bbgg
BbGg
Bbgg
bbGg
bbgg
BbGg
Bbgg
bbGg
bbgg
A
4 points ( 1 pt. each correct )
What is the phenotypic ratio?
1:1:1:1 or 4:4:4:4
Black hair / green eyes = 4
Black hair / yellow eyes = 4
brown hair / green eyes = 4
brown hair / yellow eyes = 4
B
points ( 1 pt. each correct )
What is the phenotypic ratio?
3:3:1:1 or 6:6:2:2
Black hair / green eyes = 6
Black hair / yellow eyes = 6
Brown hair / green eyes = 2
Brown hair / yellow eyes = 2