Dominance Inheritance Review
Incomplete Inheritance
Co-dominance Inheritance
SBI3U
• When one (1) allele is stronger (so dominant) than the other allele in
heterozygous genotype, resulting in phenotype of the strong (dominant)
Examples:
a) For hair texture trait  Curly hair is dominant (C)
 Straight hair is recessive (c)
 Therefore, Cc genotype translates to Curly hair phenotype
b) For dimple trait  Having dimple is dominant (D)
 Not having dimple is recessive (d)
 Therefore, Dd genotype translates to having dimple phenotype
c) For hitchhiker thumb trait  Having the hitchhiker thumb is dominant (_____)
 Not having the hitchhiker thumb is _______ (___)
 Therefore, ______ genotype translates to _______
Sample problem:
Dwarfism, is a dominant trait.
If both parents with dwarfism phenotype have normal looking offspring,
a) determine the genotype of the parents
b) the percentage of their child looking normal
c) the percentage of their grandchildren from the normal offsprings looking dwarfish
given that their life partners are also normal
Answers:
a) P: Dd x Dd (because if either one of
the parents has DD, none of their
children will look normal)
b)
D d
D DD Dd
DD =
d Dd dd
dd =
Dd =
1
4
2
4
1
4
= 25% (dwarfism)
= 50% (dwarfism)
= 25% (normal)
c) Zero percent (0%)  F1 = dd x dd
Sample problem:
Breast cancer has been tracked to associate with genes BRCA1 & BRCA2, which are
tumour suppressors and function to inhibit the growth of tumour (cell cycle genes)
It is a recessive trait. That means, if the gene is negatively mutated and the
offsprings carry both mutated alleles, she is more likely to get breast cancer.
a) Determine the phenotype of the genotype variants for this trait
b) If the mother is heterozygous for the trait and the father has normal alleles,
how many of their 6 children will
i) be carrier of the cancer trait (heterozygous)
ii) be homozygous for normal BRCA1
BRCA1 brca1
iii) develop breast cancer
BRCA1
BRCA1
BRCA1 BRCA1
brca1
Answers:
BRCA1
a) BRCA1 + BRCA1  normal; least risk for cancer
BRCA1 + brca1  carrier of the cancer trait
brca1 + brca1  develop cancer
b) i) 6
ii) 6
BRCA1
BRCA1
2
4
2
4
iii) zero
BRCA1
brca1
= 3 children
= 3 children
• Also known as Blending Inheritance
• When two (2) alleles are equally dominant (same strength), they
interact to produce a new, third, phenotype
 This means, the heterozygous genotype has its own phenotype
 Three (3) phenotypes are produced from such cross
• Although Mendel did not observe such pattern of inheritance with
peas, there are many examples found in nature.
When the
heterozygous
genotype produces
a different/new
phenotype
A red hibiscus is crossed with a white hibiscus
What are their genotypes?
P: (HR HR) x (HW HW)
F1: HRHR = ¼ = 25%  Red
HR HW = ½ = 50%  Pink
HW HW = ¼ = 25%  White
F2: Predict the phenotype ratio
for self-pollinated hybrid
hibiscus.
Your Turn:
In guinea pigs, colour of coat is determined by at least three alleles.
 Yellow coat is determined by the homozygous genotype YY,
 White by the homozygous genotype WW, and
 Cream by the heterozygous genotype YW.
Determine the expected genotype and phenotype ratio of the F1
generation which would result from a cross between:
a) two cream coloured guinea pigs;
•b) a yellow coated and a cream coated animal
… Solutions
Let GYGY represent yellow coat
Let GWGW represent white coat
Let GYGW represent cream coat
a) P generation phenotypes:
genotypes:
GY
GW
cream x cream
GYGW x GYGW
GY
GYGY GY GW
GYGY = ¼ = 25% (yellow)
GYGW = ½ = 50% (cream)
GW
GYGW GW GW
GW GW = ¼ = 25% (white)
F1 :
genotypes
Therefore, the expected genotypic ratio is: 1 : 2 : 1
the expected phenotypic ratio is: 1 : 2 : 1
b) P generation phenotypes:
genotypes:
GY
GY
GY
GYGY GY GY
GW
GYGW GYGW
cream x yellow
GYGW x GYGY
F1 :
genotypes
GYGY = ½ = 50% (yellow)
GYGW = ½ = 50% (cream)
Therefore, the expected genotypic ratio is: 1 : 1
the expected phenotypic ratio is: 1 : 1
1. Howdy! My name is Bob Howard, and I own 20 purebred red cows.
Something strange happened several months ago. During a violent storm, all
of the fences that separate my cattle from my neighbors cattle blew down.
During the time that the fences were down, three bulls, one from each
neighbor, had access to my cows. For awhile, I thought that none of the bulls
found my cows, but over the months, I have come to the conclusion that all of
my cows are expecting calves. One of the bulls is the father. Which bull is it?
A local college professor told me to use a little genetics detective work to
figure out who the father is. He told me to collect information about each of
the bulls, and to read articles about genetics and Gregor Mendel's
experiments in genetics. So, I did exactly what he said. I compiled the
information. Now, I need your help to make sense of the data and to figure
out who the father is.
After reading through the information, maybe you can tell me why my red
cows had 9 roan calves and 11 red calves. I don’t really understand how this
happened. When you have determined which bull is the father, please tell
me the answer.
This is Rocky. He is a 2,200 pound Red bull. The colour of Rocky&’s
calves, if mated with a red cow, can be determined by using a Punnett square.
His offspring will also be unique in colour compared to the other two bulls.
This is Rufus. He is a 1,920 pound White bull. The colour of Rufus’
calves can also be figured out, if he is mated with a red cow, by using a
Punnett square. The colour of his calves will also differ from the other two
bulls offspring.
This is Ferdinand. He is 2,000 pound Roan bull. The laws of genetics tell
us that the offspring he produces will probably be different, in colour, than the
other two bulls’ offspring. Using a Punnett square, you can see the gene
combinations, for colour, that Ferdinand’s offspring could have if he mates
with a red cow. The colour of Ferdinand’s calves has to do with probability.
Who did it?
… a bonus mark for first correct answer posted on Wiki discussion
Co-dominance
• When the dominancy (strength) of the allele changes unpredictably
 That is, sometimes one allele is dominant; other times it is not
 This results in the re-appearance of the both parental phenotypes
which causes the offspring phenotype to look patchy!
• The genotypes are not blended and they still obey Mendel’s law of
segregation, resulting in a mixture of the phenotype.
Haemoglobin
normal red
blood cell
The gene for haemoglobin Hb
has two co-dominant alleles:
i) HbA (the normal gene)
ii) HbS (the mutated gene)
Heterozygous: HbA HbS is good for protecting from Malaria
sickled red
blood cell
Cat coat: orange (OO) x black (oo) = Tortoiseshell
maternal
‘O’ allele
paternal
‘o’ allele
mosaic adult
Phenotype Genotype
Picture
Antibodies
Notes
Type A
IAIA or IAi
B – Antibodies
Type B
IBIB or IBi
A - Antibodies
Type AB
IAIB
No antibodies
Universal
Recipient
Type O
ii
A – Antibodies
B – Antibodies
Universal
Donor
http://gslc.genetics.utah.edu/units/basics/blood/types.cfm