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Genetics Notes Part I
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The study of genetics examines the inheritance of biological traits.
The passing of traits from parents to offspring is called heredity.
Biological traits are coded for by the genes on our chromosomes.
Half of your chromosomes came from your mother and ½ from your father.
It is estimated that more than 8 million combinations are possible from the 23 chromosome pairs you
inherited from each of your parents. That means that there are about 64 trillion possibilities of unique
combinations that create you. In short, you are uniquely different than any other individual.
Historical Development of Genetic Understanding
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An Austrian monk by the name of Gregor Mendel (1822-1884) conducted some of the
first recorded scientific studies on inheritance.
 He is currently considered the father of modern genetics. The study of inheritance is
commonly known as Mendelian genetics.
 Mendel worked with garden peas. Why? Well, he noticed that peas had very consistent
traits. Some produced green peas, others produced yellow peas. Some were tall, others
were short. Some produced flowers on the ends of the stem; others produced their
flowers on the sides of the stem. Some produced white flowers, others produced purple
flowers.
 By actively pollinating one pure plant (a plant that always produced the same
characteristics in the offspring) with a different pure plant, he could see how traits were expressed in the
offspring of the cross-pollination.
Early hypothesis suggested that if you crossed two different traits, the result would be a trait somewhere
between the two. For example, crossing a dark purple flowered plant with a white flowered plant would produce
a light purple plant. Crossing a tall plant with a short plant would produce a medium height plant. However, this
was NOT found to be the case. It appeared that one trait always dominated the other trait.
He found that tall plants produced TALL plants, even if crossed with short plants. Yellow peas produced yellow
peas even when crossed with green peas.
Although not known at the time, it was the genes that were controlling the genetic traits seen in the offspring.
Even though the offspring obtained
different possible versions of those
same genes (known as alleles) from
its two parents, only one of those
alleles would be expressed (seen).
Your mother may have blue eyes
(her chromosomes contain the
alleles for blue eye colour) and
your dad might have brown eyes
(his chromosomes contain the
alleles for brown eyes), so you
might have blue or brown eyes
(unfortunately, with eye colour
there are not only 2 possibilities, so
it does get a bit more complicated.)
In garden peas, the traits that were
expressed most often were
considered to be dominant traits.
Those that were expressed less
frequently were known as
recessive traits.
In the pea experiments, the allele for yellow seeds was dominant over the allele for green seeds. The allele for
tall plants was dominant over the allele for short plants.
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During Mendel’s experiments, he developed a short hand for expressing the traits that were seen. He
used a upper case letter to denote a dominant allele and a lower case letter to denote a recessive allele.
In his experiment, he began by cross-pollinating a pure round seed plants (identified as RR) with a
pure wrinkled seed plants (identified as rr).
The offspring were all found to be round in appearance (this is called the phenotype – which trait is
expressed/seen), but they all contained the alleles for both round and wrinkled seeds (this is called
their genotype – which alleles are present in their DNA).
These first generation offspring were called F1 generation which stands for filial 1 (filial is a Latin
word that means “son”). All of the F1 offspring contains Rr (one dominant and one recessive gene).
This makes them hybrids.
Mendel continued his experiment by crosspollinating two of the hybrids (F2). Here is what he
found:
o Three of the seeds were round and one was
wrinkled (phenotypes). ¾ : ¼
o However, looking at their genotypes:
 1 was homozygous round (RR),
 2 were heterozygous round (Rr) and
 1 was homozygous wrinkled (rr).
 1:2:1
In order to express the recessive allele, the offspring must be homozygous (possess only recessive
alleles) for that trait.
If a dominant allele is present it will mask the presence of the recessive allele. This is known as the
Principle of Dominance.
During the formation of sex cells, the homologous alleles separate and recombine. This is known as
the Law of Segregation.
Try This Activity
Creating a Personal Profile
Table #1: Lists of human traits controlled by dominant and recessive alleles.
Trait
Dominant
Eye colour
Brown or black or green
Hair colour
Brown or black
Hairline
Pointed on forehead
Freckles
Present
Earlobe
Suspended
Hair texture
Curly
Eyesight
Near or far sighted
Eyelashes
Long
Nose line
Convex tip
Fingers
6 fingers
Rh blood factor
Positive Rh factor
Ear rim
Curled rim
Thumb joint
Last joint bends out
Finger hair
Present
Folded Hands
Left thumb over right
Tongue Rolling
Can be rolled into U shape
Clenched Fist
Two wrist cords
Chin Dimple
Dimple in middle
Blood type
Type A, B and AB
Eyes
Astigmatism
Recessive
Blue or grey
Blonde or red
Straight across forehead
Absent
Attached to head
Straight
Normal vision
Short
Concave or straight
5 fingers
Negative Rh factor
Not curled rim
Last joint straight
Absent
Right thumb over left
Cannot be rolled
Three wrist cords
No dimple
Type O
No astigmatism
Table #2: Use information from Table 1 to complete your personal profile. Which additional traits can you
include?
Trait (use the letter
Appearance or physical
Possible genetic makeup
Dominant or recessive
indicated)
condition (phenotype)
(genotype)
Eye colour
E/e
Hairline
L/l
Earlobe
T/t
Ear rim
R/r
Freckles
F/f
Thumb joint
J/j
Finger hair
P/p
Tongue rolling Y/y
Folded hands
D/d
Nose line
N/n
Hair colour
H/h
Chin dimple
G/g
Clenched fist
K/k
Single-Trait Inheritance
Define each of the following:
1. Genotype: ___________________________________________
_____________________________________________________
2. Phenotype:___________________________________________
_____________________________________________________
3. Homozygous:
_______________________________________
_____________________________________________________
4. Heterozygous:
_______________________________________
_____________________________________________________
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The cross between a TT tall pea plant and a tt short pea plant is called a
monohybrid cross. Crossing a Tt tall pea plant with another Tt tall pea plant is also
a monohybrid cross. A monohybrid cross is a cross that involves one allele pair of
contrasting traits.
A special chart, called a Punnett square, helps us organize the results of a cross
between the sex cells of two individuals.
Problem #1
Consider a cross between a pea plant that is heterozygous for round seeds and a pea plant that has
wrinkled seeds. The allele for round seeds is dominant over that for wrinkled seeds. R can be used to
indicate the round dominant allele and r can be used to represent the wrinkled recessive allele.
A)
Determine the genotypes of the offspring.
B)
Determine the phenotypes of the offspring.
Wrinkled Parent (rr)
Punnett Square
r
R
Round parent
(Rr)
r
r
Problem #2
Consider a cross between two pea plants that are heterozygous for round seeds. The allele for round seeds
is dominant over that for wrinkled seeds. R can be used to indicate the round dominant allele and r can be
used to represent the wrinkled recessive allele.
A)
Determine the genotypes of the offspring.
B)
Determine the phenotypes of the offspring.
Round Parent (Rr)
Punnett Square
R
r
R
Round parent
(Rr)
r
Problem #3
Consider a cross where only the offspring are observable. It is still possible to determine the genotypes of
the parents in many cases even if the parents are unknown.
Offspring phenotype
Round seed peas
Wrinkled-seed peas
A)
Numbers
5472
1850
Determine the ratio of round plants to wrinkled plants. You may need to round the ratio.
round
 _________  _____
wrinkled
B)
Now list the possible genotypes for each phenotype:
Offspring phenotype
Round seed peas
Wrinkled-seed peas
C)
Genotype Possibilities
RR or Rr
rr
Based on your results from the last two questions, what phenotype did the parents of these
results have?
Problem #4
A plant that is homozygous for purple flowers is crossed with a plant that has white flowers. If purple is
dominant over the white condition, what are the genotypes and phenotypes of the F1 generation?
White Parent (pp)
Punnett Square
Purple Parent
(PP)
Problem #5
Table #1: Results of Mendel’s Experiments
Trait
Plant height
Flower colour
Flower position on stem
Pod colour
Pod shape
Seed colour
Seed shape
Alleles
DOMINANT
recessive
Possible genotypes
TALL
dwarf
PURPLE
white
AXIAL (at branches)
top
GREEN
yellow
INFLATED
constricted
YELLOW
green
ROUND
wrinkled
TT or Tt
tt
PP or Pp
Pp
AA or Aa
Aa
GG or Gg
Gg
II or Ii
Ii
YY or Yy
Yy
RR or Rr
rr
For each of the crosses listed below, create a Punnett square and determine the following information:
parent phenotypes, parent genotypes, parent gametes, F1 genotypes and F1 phenotypes.
a)
Two heterozygous tall parents are crossed.
Parent phenotypes
Punnett Square
Parent genotypes
Parent gametes
F1 genotypes
F1 phenotypes
b)
A heterozygous tall plant is crossed with a dwarf plant.
Punnett Square
Parent
phenotypes
Parent genotypes
Parent gametes
F1 genotypes
F1 phenotypes
c)
Two plants that are heterozygous for purple flowers are crossed.
Punnett Square
Parent
phenotypes
Parent genotypes
Parent gametes
F1 genotypes
F1 phenotypes
d)
A plant that is homozygous for green pods is crossed with a plant that has yellow pods.
Punnett Square
Parent
phenotypes
Parent genotypes
Parent gametes
F1 genotypes
F1 phenotypes
e)
A plant that is homozygous for round seeds is crossed with a plant that is heterozygous for
round seeds.
Punnett Square
Parent
phenotypes
Parent genotypes
Parent gametes
F1 genotypes
F1 phenotypes
Problem #6
In guinea pigs, the allele for a black coat is dominant over the allele for a white coat. A black guinea pig
was crossed with a white guinea pig. All F1 offspring have black coats.
A)
Describe how you can determine whether or not the black parent is homozygous or heterozygous
for the black condition. Indicate the letter you will use to represent an allele
Punnett Square
b
b
B
BB
Bb
B
BB
Bb
b
b
B
BB
Bb
b
Bb
bb
Punnett Square
B)
If 10 offspring were produced, indicate how many you would expect to have black coat
colour, if the black parent were heterozygous.
Punnett Square
b
b
B
Bb
Bb
b
bb
bb
Test Cross
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Sometimes in agriculture a particular trait is desired. For example, white sheep are preferred over
black sheep because the wool of black sheep is more brittle and harder to dye. Therefore, farmers
strive to selectively breed their flock to produce only white sheep. How do they ensure the offspring of
a mating will be white? Let’s look at the genetics behind the alleles that code for white or black wool.
White wool is the dominant condition and black wool is the recessive condition.
To ensure the offspring are all white, ranchers will use a
homozygous white ram (male sheep). How do they know the
ram is homozygous? Ranchers perform a test cross.
The test cross is always performed between the unknown
genotype and a homozygous recessive genotype. In this case,
the recessive individual would be a black ewe (female sheep).
In a cross between a white ram whose genotype is unknown
(either WW or Ww) and a black ewe (ww), if any of the
offspring show the recessive trait, then the ram must be
heterozygous. If all the offspring show the dominant trait, then
the ram must be homozygous.
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Punnett Square
White
Ram
Unknown
genotype
Homozygous?
W
W
Black ewe
(recessive)
w
w
Punnett Square
Ww
Ww
(white)
(white)
White
Ram
Ww
Ww
(white)
(white)
Unknown
genotype
Heterozygous?
W
w
Black ewe
(recessive)
w
w
Ww
Ww
(white)
(white)
ww
ww
(black)
(black)
Problem #7
For Labrador Retrievers, black fur (F) colour is dominant over yellow (f).
A)
What would be the genotype of a homozygous black dog? ________________________________
B)
Could a heterozygous black dog have the same genotype as a dog with yellow fur? Explain.
Problem #8
A heterozygous pea plant with round seeds (R) is cross-pollinated with a pea plant that has wrinkled seeds
(r). Remember that the round condition is dominant over the wrinkled seed. For the cross, indicate each of
the following:
A)
What would be the genotype of the parents if the round-seed plant is heterozygous?____________
B)
What are the genotypes and phenotypes of the F1 generation?
Punnett Square
F1
Wrinkled
r
r
R
Rr
Rr
r
rr
rr
Round
C)
What are the genotypes and phenotypes of the F2 generation, if two round-seed plants from F1 are
cross-pollinated?
Punnett Square
F2
R
r
R
RR
Rr
r
Rr
rr
Problem #9
For Dalmatian dogs, the spotted condition (S) is dominant over non-spotted (s).
A)
Using a Punnett square, show the results of a cross between two heterozygous parents.
Non-Spotted
Punnett Square
F1
Spotted
B)
A spotted female Dalmatian mates with an unknown father. From the appearance of the pups, the
owner concludes that the male was a Dalmatian. The owner notes that the female had six pups,
three spotted and three non-spotted. What are the genotype and phenotype of the unknown male?
Male's genotype:
Spotted Female
Punnett Square
F1
S
s
_______________
s
Ss
ss
Male's phenotype:
s
Ss
ss
_______________
Unknown
Male?
Problem #10
A human neurological disorder referred to as Huntington's chorea is caused by a dominant allele. Because
the allele doesn't express itself until a person reaches 50 years of age, early detection has been difficult. In
one family, a woman begins to show symptoms. Her father had Huntington's chorea, but her mother never
developed the disorder. The woman's husband shows no symptoms, nor does anyone in his immediate
family.
A) What is the genotype of the woman who has developed Huntington's chorea?
B) What is the probable genotype of the woman's husband?
C) If the man and woman have 6 children together, how many are likely to develop Huntington's chorea?
Woman
Punnett Square
F1
H
h
h
Hh
hh
h
Hh
hh
Man
Selective Breeding
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People have been using selective breeding for a long time as a way of improving varieties of plants
and animals. Selective breeding involves the crossing of desired traits from plants or animals to
produce offspring that have one or several of the favoured characteristics.
For example, farmers might selectively cross-pollinate a cold-hardy wheat with one that grows quicker
in order to create a new variety that can survive in Canada's colder, shorter growing season.
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Purebred pets and thoroughbred horses have been selectively bred over many generations using
inbreeding. Inbreeding is the mating of closely related individuals for the purpose of maintaining or
perpetuating certain characteristics.
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New varieties of plants and animals can be developed by hybridization. This process is the opposite
of inbreeding. In hybridization, the goal is to combine different traits from different parents to have all
of these favorable traits present in the offspring. Hybrids tend to be more vigorous than either parent.
Standard poodle
Labrador retriever
Labradoodle
Pedigree Charts
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A pedigree chart can be used to solve genetic puzzles when traits have been recorded over many
generations.
A pedigree shows the passing of a trait through the generations.
Pedigree charts use many symbols to represent gender, relationships between individuals, and whether
or not the individual expresses a trait or carries the allele as part of a heterozygous genotype.
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hjg
Problem #11
A) Indicate whether each family member is
homozygous or heterozygous for
shortsigthedness or homozygous for normal
vision.
I-1
ss
II-4
Ss
I-2
Ss
II-5
Ss
II-1
Ss
III-1
Ss
II-2
ss
III-2
Ss
II-3
ss
III-3
ss
I
2
1
II
1
2
4
3
5
III
1
2
?
4
3
This pedigree chart shows a family with the trait of shortsightedness.
The allele for shortsightedness (S) is dominant over the allele for
normal vision (s)
B) If couple II - 4 & 5 (in row II) had another child, what genotype
might the child have? (Hint: What genotype is possible but not
shown in the chart? Would the child have normal vision or be
shortsighted?
Punnett Square
F1
Woman
S
s
S
Ss
Ss
s
Ss
ss
Man
Pedigree Assignment
Name: ____________________________
Section: ___
Diabetes is a recessive genetic disorder. A defective
gene reduces insulin production by the pancreas.
Insulin is released into the circulatory system and
allows the cells of the body to absorb glucose from
the blood. Individuals who lack insulin have high
blood sugar. In an attempt to trace the inheritance of
the defective allele in one family, the data in the
chart was gathered.
Date: ______________
Name
Relationship
Jennifer
Ryan
Helen
James
Walter
Susan
Colin
Mother
Father
Daughter
Son
Son
Wife of Walter
Son of Walter & Susan
A) Construct a pedigree chart showing the passage of the diabetes allele.
B) Indicate the probable genotypes of Jennifer and Ryan.
C) Indicate the probable genotypes of Susan and Walter.
D)
Whose genotype cannot be determined with 100% certainty? Explain.
Phenotype
Normal (N)
Normal (N)
Normal (N)
Normal (N)
Diabetic (n)
Normal (N)
Diabetic (n)
Multiple Alleles
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When Mendel did his experiments, there were only 2 possible alleles for each trait. However, many
traits have more than just two alleles (e.g. eye colour can be brown, blue, green, grey, hazel, etc.).
When there are multiple alleles, we no longer use upper and lower case letters. Instead, we use
superscripts. For example, in human blood typing, blood type A, if homozygous, contains 2 alleles,
IaIa.
Geneticists who study fruit flies (Drosophila melanogaster) noted that there are many different eye
colour possibilities. The most common eye colour is red (called Wild), but they can also have apricot,
honey and white eyes.
In fruit flies there is a dominance hierarchy (this means that each colour is successfully dominant over
the colour below it on the list). The hierarchy is:
Phenotype
Possible Genotype
Dominant over
1
1 1
1 2
1 3
1 4
Wild type (E )
EE, EE, EE, EE
Apricot, honey, white
Apricot (E2)
E2E2, E2E3, E2E4
Honey, white
3
3 3
3 4
Honey (E )
EE, EE
White
4
4 4
White (E )
EE
none
Note that with the fruit fly dominance hierarchy, E is used to represent "eye colour" and the
superscript represents their order in the hierarchy.
Problem #12
What is the phenotype ratio of the offspring from the mating of the following Drosophila?
A)
Note:
Always write the
more dominant allele
(lower number) for
the offspring 1st.
E1E4 (wild-type eye colour) x E2E3 (apricot eye colour)
Punnett Square
F1
Apricot
E2
E3
E1
Wild
E4
B)
E3E4 (honey colour) x E4E4 (white eye colour)
Punnett Square
F1
E3
Honey
E4
White
4
E
E4
Problem #13
Using the dominance hierarchy chart for Drosophila, answer the following questions?
A)
Of the genotypes listed, which would represent the homozygous condition? Explain.
B)
Of the genotypes listed, which would represent the homozygous recessive condition? Explain.
C)
Find the genotypes and the phenotypes (F1) if a white-eyed fruit fly is crossed with one that has
honey-coloured eyes.
Punnett Square
F1
Honey-coloured
E3
E3
E3
E4
E4
White
E4
D)
Find the phenotypes if a fly with apricot eyes (E2E4 ) is crossed with one that has honeycoloured eyes (E3E4)
Punnett Square
F1
Honey-coloured
apricot
D)
Find the phenotypes if a fly with wild eyes (E1E3 ) is crossed with one that has honeycoloured eyes (E3E4)
Punnett Square
F1
wild
Honey-coloured
Incomplete Dominance and Codominance
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In Mendel's day, he only encountered species that always had a dominant trait and recessive traits.
However, there are species where the alleles are equally dominant. When they interact, they produce a
new phenotype that is a blended trait. This is called incomplete dominance.
When a red snapdragon is crossed with a white snapdragon, all of the F1 offspring are pink. This type
of incomplete dominance is called intermediate inheritance.
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Punnett
Square
F1
CW
White
CW
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CW
Pink
CR
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CR
CR
CW CR
pink
CW CR
pink
CW CR
pink
CW CR
pink
If the F1 offspring are allowed to self-fertilize, the F2
generation will produce a ratio of one red, to two pink,
to one white.
Punnett
Square
F2
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Red
Pink
CW
CR
CW CW
white
CW CR
pink
CW CR
pink
CR CR
red
Another type of incomplete dominance is referred to as codominance.
In this type of interaction, both alleles are expressed at the same time. For example, short horn cattle
show this type of incomplete dominance.
When a red bull is crossed with a white cow, the offspring will be a roan calf (a calf with red and
white hair). Roan calves are also produced when a white bull is crossed with a red cow.
Problem #14
A)
Define:
i)
Dominance:
__________________________________________________________
______________________________________________________________________
ii)
Codominance: __________________________________________________________
______________________________________________________________________
iii)
Incomplete dominance:
______________________________________________
______________________________________________________________________
B)
Determine the F1 phenotypes of a cross between a pink and a white snapdragon.
Punnett
Square
F1
Pink
CW
CR
CW
White
CW
C)
Find the F1 phenotypes of a cross between a white cow and a roan bull.
Punnett
Square
F1
White
Cow
D)
Roan Bull
CR
CW
CW
CW
A geneticist notes that crossing a round radish with a long radish produces oval radishes. If
oval radishes are crossed with oval radishes the following phenotypes of notes: 100 long, 200
oval & 100 round radishes. Create Punnett squares for the 2 generations to explain the results.
Punnett
Square
F1
Punnett Square
F2
A Murder Mystery Case Study
Name: ______________________
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Section: ____
Date: ______________
There are 4 different ABO blood types as shown in the table here.
The alleles for blood types A and B are codominant but are both
dominant over the allele for type O.
The rhesus factor is a blood factor that is regulated by a gene. The
Rh-positive allele is dominant over the Rh-negative allele. In this
case study, you will solve a murder mystery using genetics.
Phenotypes
Genotypes
Type A
IAIA, IAIO
Type B
IBIB, IBIO
Type AB
IAIB
Type O
IOIO
Evidence:
As a bolt of lightning flashed above Black Mourning Castle, a scream echoed from the den of Lord
Hooke. When the upstairs maid peered through the door, a freckled arm reached for her neck. Quickly,
the maid bolted from the doorway, locked herself in
the library, and telephoned the police. Inspector
Holmes arrived to find a frightened maid and the
Lord Hooke
Lady Hooke
dead body of Lord Hooke. Apparently, the Lord had
been strangled. The inspector quickly gathered
evidence. He noted blood on a letter opener, even
Tom
Jane
Ann
Ida
Helen
though Lord Hooke did not have any cuts or
Roule
abrasions. The blood sample proved to be type O,
Rh-negative. The quick thinking inspector phoned
the family doctor for each family member's medical
history. The following shows the relatives who were
Henry
Tina
Beth
in the castle at the time of Lord Hooke's murder.
The inspector gathered the information in the
following table. Some of the family members were
deeply tanned, so the inspector found it difficult to
determine whether or not freckles were present on
their arms. Note that having freckles is an inherited
trait and the allele for freckles is dominant over the
allele for no freckles.
The crafty inspector drew the family close together
and, while puffing on his pipe, indicated that he had
found the murderer. He explained that one of the
heirs to the fortune was not Lord Hooke's biological
child. The inspector believed that the child
committed the murder.
Family
Member
Lord Hooke
Lady Hooke
Helen
Roule
Henry
Ida
Ann
Tom
Jane
Beth
Tina
Analysis
A)
B)
C)
Who was the murderer? State the reasons for your answer.
Describe the procedure you followed to obtain your answer.
How did the inspector eliminate the other family members?
Blood
Type
AB
A
A
O
Rh
Factor
+
+
+
+
Refused
blood test
A
B
O
A
O
A
+
+
+
Freckles
No
No
No
No
?
?
?
No
?
?
Yes
Dihybrid Crosses
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Mendel also experimented with two traits at the same time.
He cross-pollinated pure-breeding plants that produced yellow, round seeds with pure-breeding
plants that produced green, wrinkled seeds.
The Laws of genetics that apply for single-trait inheritance (monohybrid cross) also apply for twotrait inheritance.
Yellow Round
As seen in this Punnett Square, yellow round plants are
Punnett
Square
YYRR
denoted with YYRR, and green wrinkled peas are denoted
F
1
as yyrr, because yellow and round are dominant traits and
YR
YR
green and wrinkled are recessive traits.
The F1 offspring from this cross are heterozygous for both
yr
YyRr
YyRr
Green
yellow and round genotypes.
Wrinkled
yyrr
yr
YyRr
YyRr
Now consider a cross between a pure-breeding yellow,
wrinkled pea plant with a green, round pea. This Punnett
Square shows the results.
Mendel found that the alleles assorted independently.
This is now known as the Law of Independent
Assortment.
Let's take a look as see what
happens if the F1 generation
from the last Punnett square
self-fertilized to produce an
F2. generation.
Punnett
Square
F2
YR
yR

Notice the outcome ratios:
Punnett Square
F1
Green,
Round
yyRR
Yellow
Wrinkled
YYrr
Yr
Yr
yR
YyRr
YyRr
yR
YyRr
YyRr
YR
yR
Yr
yr
YYRR
YyRR
YYRr
YyRr
Yellow
Round
Yellow
Round
Yellow
Round
Yellow
Round
YyRR
yyRR
YyRr
yyRr
Yellow
Round
Green
Round
Yellow
Round
Green
Round
YYRr
YyRr
YYrr
Yyrr
Yr
Yellow
Round
Yellow
Round
Yellow
Wrinkled
Yellow
Wrinkled
YyRr
yyRr
Yyrr
yyrr
yr
Yellow
Round
Green
Round
Yellow
Wrinkled
Green
Wrinkled
Yellow,
Round
Yellow,
Wrinkled
9/16
3/16
Green,
Round
Green,
Wrinkled
3/16
1/16
Probability of Outcomes

Genotypic and phenotypic ratios are determined by the probability of inheriting a certain trait.
Probability =



# of ways that a given event could occur
total number of possible events
For example, in a coin toss, there is only one way of tossing heads, so the numerator is 1. The denominator is
2 because there are two possible events in total (heads or tails).
Rule #1 of Probability: Chance has no memory –
o If you tossed two heads in a row, the probability of tossing heads once again would still be ½ . The
probability of each event is not affected by the results of the other events. They are all independent.
Rule #2 of Probability: The Probability of independent events occurring together is equal to the products of
those events occurring separately.
o For example, the chances of tossing heads once is ½; the probability of tossing heads twice is ½ x ½
= ¼ and the probability of tossing heads three times in a row is ½ x ½ x ½ = 1/8
Sample Problem 1
In humans, free earlobes are controlled by the dominant allele E, and attached earlobes by the recessive allele e.
The widow’s peak hairline is regulated by the dominant allele H; while the straight hairline is controlled by the
recessive allele h. Consider the mating of the following genotypes:
EeHh x EeHh
What are the probabilities of obtaining F1 offspring with the following characteristics?
1)
2)
3)
4)
Widow’s peak with free earlobes
Straight hairline with free earlobes
Widow’s peak and attached earlobes
Straight hairline and attached earlobes
Solution
Dihybrids can be treated as two monohybrids. Isolate the gene for earlobes and for hairline and work with each
as a monohybrid. The F1 generation resulting from a cross between the heterozygous parents can be determined.
E
e
E
EE
Ee
e
Ee
ee
Free earlobes ¾
Widow’s peak ¾
H
h
H
HH
Hh
h
Hh
hh
Attached earlobes ¼
Straight hairline ¼
Punnett
Square
F1
EH
eH
Eh
eh
EH
EEHH
EeHH
EEHh
EeHh
eH
EeHH
eeHH
EeHh
eeHh
Eh
EEHh
EeHh
EEhh
Eehh
eh
EeHh
eeHH
Eehh
eehh
The monohybrids can not be combined to calculate the probabilities of the dihybrid crosses.
1)
2)
3)
4)
Widow’s peak with free earlobes
Straight hairline with free earlobes
Widow’s peak and attached earlobes
Straight hairline and attached earlobes
¾ x ¾ or 9/16
¼ x ¾ or 3/16
¾ x ¼ or 3/16
¼ x ¼ or 1/16
Problem 15
What is the probability that child from the mating of the EeHh x EeHh parents for the following:
1)
The probability that the child would be a male? __________________________________
2)
The probability of having a widow’s peak? ______________________________________
3)
The probability of having attached earlobes? _____________________________________
4)
The probability of the child beings a male, having a widow’s peak and attached earlobes?
___________________________________________
At this point, complete the questions on Page 158 -159 in the Nelson Biology 11 textbook.