Human GeneticsHands-on labs, inc. Version 42-0059-00-01.docx

Human Genetics
Hands-on labs, inc. Version 42-0059-00-01
PHOTOS: There are no photos to be included with
this lab report.
Exercise 1: genotype to Phenotype
Observations
Experiment
Human Genetics
281
©Hands-On Labs, Inc.
www.LabPaq.com
Data Table 1: Trait Characteristics for Exercise 1
#
Trait
Allele
from
Mother
Allele
from
Father

Child’s
genotype
Child’s
Phenotype
(written)
1
Face shape
● AA, Aa aa
● Round Square
A or a
A or a
2
Chin size
● BB, Bb bb
● ProminentAverage
B or b

B or b
3
Cleft chin
● CC, Cc cc
● Cleft chin No cleft
C or c
C or c
4
Hair type
● DD Dd dd
● Straight Wavy Curly
D or d
D or d

5
Widow’s peak
● EE, Ee ee
● Widows peak Absent
E or e
E or e
6
White forelock
● FF, Ff ff
● White forelock Absent
F or f
F or f
7
Eye shape
● GG, Gg gg
● Almond shape Round shape
G or g
G or g

8
Eye slantedness
● HH, Hh hh
● Horizontal Upward slant
H or h
H or h
9
Eyelashes
● II, Ii ii
● Long lashes Short lashes
I or i
I or i
10
Eyebrow thickness

● JJ, Jjjj
● Thick Thin
J or j
J or j
11
Eyebrow length
● KK, Kk kk
● Separated brows Brows joined
K or k
K or k
12
Lip thickness
● LL, Ll ll
● Thick lips Thin lips
L or l
L or l

13
Dimples
● MM, Mm mm
● Dimples No dimples
M or m
M or m
14
Nose shape
● NN, Nn nn
● Round Pointed
N or n
N or n
15
Nostril shape
● OO, Oo oo
● Round Pointed

O or o
O or o
16
Earlobe attachment
● PP, Pp pp
● Free Attached
P or p
P or p
17
Darwin’s earpoint
● QQ, Qq qq
● Pointed Point absent
Q or q
Q or q
18
Cheek freckles

● SS, Ss ss
● Present Absent
S or s
S or s
19
Forehead freckles
● TT, Tt tt
● Forehead freckles Absent
T or t
T or t
Observations
A. Why was the male chosen to determine the sex of the child?

B. Draw a picture of the compiled traits.
C. If this activity was repeated two or three times would the
offspring images change? Explain
your answer.
Exercise 2: Punnett Squares
Questions
Use a Punnett square to solve the following scenarios. You

MUST INCLUDE THE PUNNET SQUARE AS PART OF YOUR
ANSWER FOR EACH OF THESE PROBLEMS.
A. Dominant: cheek freckles
Recessive: no cheek freckles
(Please note that this trait is more complicated than described in
this lab exercise, but for learning purposes, assume that this is
correct.)
A man with a heterozygous gene for freckles on his cheeks has a
child with a woman who does not have freckles on her cheeks.
What is the likelihood that their son will have freckles on his
cheeks?
B. Dominant: normal; no cystic fibrosis
Recessive: cystic fibrosis

A woman has a history of cystic fibrosis in her family, and
found out that she has the gene for cystic fibrosis, but is not
affected by the gene. Her husband also has a history of cystic
fibrosis in his family. He got tested yesterday and found that he
is also a carrier for the disease, but is not affected by it. They
are going to have a baby in two months. What is the likelihood
that their baby will have cystic fibrosis?
C. Dominant: normal; no Tay-Sachs
Recessive: Tay-Sachs
A man is a carrier for Tay-Sachs. He is going to have a child
with a woman who has homozygous normal genes. What are the
chances that they will have a child who has Tay-Sachs disease?
D. Dominant: Round face Dominant: Prominent chin

Recessive: Square face Recessive: Non-prominent chin
(Please note that this trait is more complicated than described in
this lab exercise, but for learning purposes, assume that this is
correct.)
A woman with heterozygous genes for a round face shape and a
chin that is not prominent has a child with a man who has a
square face and homozygous genes for a prominent chin. What
are the chances that the child has a square face with a prominent
chin?
E. Dominant: Huntington’s disease Dominant:
normal; no cystic fibrosis
Recessive: normal; no Huntington’s Recessive: cystic
fibrosis
A man with Huntington’s disease is also a carrier for cystic
fibrosis. His wife has cystic fibrosis but does not have
Huntington’s disease. If they have a baby, what is the likelihood
that the baby will have both Huntington’s disease and cystic
fibrosis?

F. X-linked dominant trait: no color blindness
X-linked recessive trait: color blindness
A color-blind man marries a non-color-blind woman who is not
a carrier for color blindness. What are the chances that they will
have a son who is also color-blind?
What are the chances that they will have a daughter who is a
carrier for color blindness?
Omit pedigrees.
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Human Gen
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42
-
0059
-
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PHOTOS: There are no photos to be included with this lab
report.
E
x
e
r
cise 1:
g
enotype
t
o
Phenotype
Observations

Experiment
Human Genetics
28
1
©Hands-On Labs, Inc.
www.LabPaq.com
Human Genetics
Hands-on labs, inc. Version 42-
0059-00-01
PHOTOS: There are no photos to be included with this lab
report.
Observations
HUMAN GENETICS Lab
Below is a listing which will help you to prepare for the quiz
and lab on this material.
CONTENT TO KNOW:
I.Review from Previous Lab:

• GENOTYPE = the genetic make-up of the individual
• PHENOTYPE = the physical manifestation of the genotype
(what the organism shows)
--note: letters are used to represent gene pairs
o Dominant allele - A gene which can hide or mask the second
gene in a pair
o Recessive allele - gene which can be masked
Ex: W - dominant - widow's peak
w - recessive - straight hairline
3 possible Genotypes:
WW - homozygous dominant
Ww - heterozygous
ww - homozygous recessive
Monohybrid Crosses -
Genotypes of the parents:

When you cross a homozygous dominant (WW) with a
homozygous recessive (ww), the results are always:
genotypic ratio - 100% Ww
phenotypic ratio - 100% dominant phenotype
(widow's peak)
Cross two parents that are heterozygous for hairline:
II. Chromosome Variations In Humans
Autosomal Recessive Inheritance –
1. Defective Gene is carried recessively
2. Males or females carry the disorder on a non-sex
chromosome (autosome)
3. Ex: Galactosemia
4. Make a genotype/phenotype table for all problems:
3 possible genotypes for Galactosemia:

Genotype Phenotype
AA Normal
Aa* Carrier
a*a* Galactosemic
Note: (*) represents the defective allele
Problem:
Cross a normal homozygous dominant parent with a carrier for
Galactosemia
Cross AA x Aa*
A A
A AA AA
a* Aa* Aa*
Results: 50% genetically normal; 50% carriers; 100% APPEAR
normal
Autosomal Dominant Inheritance
1. Males or females carry the defect on a non-sex chromosome
(on an autosome)
2. Trait is dominant
3. Example: Huntington's Disorder - a progressive degeneration
of the nervous system
(onset > 40 yrs)
4. Example: Achondroplasia (dwarfism)
5. Make a table for Autosomal Dominant Disorders

Genotype Phenotype
A*A* Huntington’s
A*a Huntington’s
aa Normal
Note: (*) indicates the defective gene
Problem : Cross a normal parent with a homozygous dominant
individual with Huntington's.
aa x A*A*
a a
A* A*a A*a
A* A*a A*a
How many of the offspring will die from Huntington's Disorder?
100%
III. Sex Chromosomes - determine the sex of the fetus
XX - 2 X chromosomes = FEMALE
Xy - one X, 1 y chromosome = MALE
Sex Determination
Mother:
XX
Possible gametes from mom: X, X

Father:
Xy
Possible gametes from dad: X, y
Sex determination
50 % male
50% female
X X
X XX XX
y Xy Xy
X-linked recessive Inheritance –
1. The mutated gene is on the X chromosome (not found on the
y)
2. Carried recessively, so a normal X can mask it
3. Ex. Hemophilia - caused by a gene for clotting factor VIII
which is mutated (causes
uncontrolled bleeding)

Genotype/Phenotype Table: (Must show both males and
females)
Xy Normal
X*y Hemophiliac
Females
XX Normal
XX* Carrier
X*X* Hemophiliac
Problem: Cross a carrier female with a normal male.
XX* x Xy
X X*
X XX XX*
y Xy X*y
Results:
25% of ALL offspring are hemophilicas
50% of FEMALES are carriers
50% of MALES are hemophiliacs
Human Genetics
Hands-on labs, inc.
Version 42-0059-00-01

Review the safety materials and wear goggles when
working with chemicals. Read the entire exercise
before you begin. Take time to organize the materials
you will need and set aside a safe work space in
which to complete the exercise.
Experiment Summary:
Students will learn how to use Punnett squares and
pedigree charts. Students will go beyond classical
Mendelian genetics to learn about incomplete
dominance, sex-linked traits, and linkage groups.
© Hands-On Labs, Inc. www.LabPaq.com 270
ExpErimEnt
ObjEctivEs
● To observe how genotypes influence phenotypes
● To predict a phenotype based on genotype information
● To use Punnett squares to predict mathematical probabilities
of offspring
● To use pedigree charts to observe genetic trends in human
lineages
TimeAllocation: 2 Hours
www.LabPaq.com 271 ©Hands-On Labs, Inc.
Experiment Human Genetics

matErials
MATERiAlS
FRoM:
lABEl oR BoX/
BAg: QTy iTEM DESCRiPTioN:
Student Provides 2 Coins
1 Pencil
1 Drawing paper, 1 sheet
1 Set of colored pencils
1 Partner
Note: The packaging and/or materials in this LabPaq may differ
slightly from that which is listed
above. For an exact listing of materials, refer to the Contents
List form included in the LabPaq.
DiscussiOn anD rEviEw
In the late 1800s, Gregor Mendel, the father of the study of
genetics, began experimenting with
traits of sweet peas. He knew that traits were inherited across
generations and was interested
in seeing if there was a pattern. His background in physics
provided the math needed to test his
ideas. His investigations showed that “units of heredity” would

separate as each parent went
through the process of forming sex cells. These units would
recombine in the offspring resulting
in the unit of heredity passing from the parent to the offspring.
This separation is called the law
of independent assortment.
Figure 1: Gregor Mendel, the father of genetics
Courtesy of NIH
Today, these “units of heredity” are called genes. Genes are
located on chromosomes. Each
person has 23 pairs of chromosomes (for a total of 46
chromosomes) present in every cell. Each
gene carries the information for the production of proteins.
Genes come in alternative forms
known as alleles. In simple Mendelian genetics, two different
alleles are present, and one allele
each is inherited from each parent. One of the alleles is
expressed or “turned on” while the other
is repressed or “turned off”. The expression of some allele pairs
can be complicated, but in this
exercise, only simple Mendelian genetics will be studied. The
expressed allele is called dominant
and the repressed allele is called recessive. The expressed trait
is referred to as the phenotype.
Each set of chromosomes are inherited from parents and align
with each other as homologous
pairs during meiosis. These pairs code for the same proteins or
genetic information. The genome
is the complete set of all the genes in an organism. genotype
refers to the genetic composition
of an organism; either as a group of genes or for individual
genetic traits.

Dominant and Recessive Alleles
Alleles are usually written as follows: A capital letter signifies
a dominant allele, such as “E” for
free (unattached) earlobes. A lowercase letter signifies a
recessive allele, such as “e” for attached
earlobes.
A person’s genotype for a specific trait can be composed of two
dominant alleles, two recessive
alleles, or one dominant allele and one recessive allele (EE, ee,
or Ee). The EE and ee genotypes
are homozygous, or the same, as both alleles are either
dominant or recessive. The Ee genotype
is referred to as heterozygous as there are two different alleles,
one for dominant and one for
recessive. In these cases, the E will be expressed as free
earlobes because it is the dominant
trait. The ee is referred to as homozygous recessive; this is the
only genotype which will produce
the recessive trait, because there are no dominant alleles to
mask the presence of this trait. The
phenotype is the physical appearance of the expression of the
alleles such as free earlobes. When
viewing the phenotype, you cannot know if the genotype is EE
or Ee. You can only know that E is
present. This example is summarized in Table 1.
Table 1: Alleles with associated genotypes and phenotypes

Allele Genotype Phenotype
E EE free earlobes
Ee free earlobes
e ee attached earlobes
Variations in Allele Expression in Humans
incomplete dominance
This state occurs when both alleles express themselves. Neither
allele is dominant nor recessive.
An example taken from this exercise is the shape of human hair:
curly, wavy or straight. The
genotypes for this trait are as follows: D = straight, Dd = wavy,
and dd = straight. However, please
note that in the case of human hair, this trait is even more
complicated than how it is described
in this lab exercise.
X-linked
These are genes located on the X chromosome (which is one of
the two sex chromosomes).
Women have two X chromosomes (their sex chromosome
genotype is XX), while men have one
X and one Y chromosome (their sex chromosome genotype is
XY). The Y chromosome is actually
“missing” an arm of a chromosome, so there are no alleles to
match with the X. The result is that
any allele present on the X chromosome will be expressed in a
man regardless of whether it is
normally recessive or dominant.

In the following activity, you will explore the probabilities
behind the hereditary process. Using
the traits in Data Table 1, you will flip coins to determine the
phenotype of offspring. A partner
will be needed for this activity.
prOcEDurE
1. Ask another person to help you with this activity.
2. Determine which person will toss for the female and which
will toss for the male. There are
two alleles per trait, and each parent will contribute one allele.
3. The person who is representing the male will flip a coin to
determine the sex of the offspring.
A heads-up toss will yield a female offspring; a tails-up toss
will yield a male offspring.
4. For all future coin tosses, heads will represent the dominant
allele and tails will represent the
recessive allele. We will assume that each parent is
heterozygous for each trait and therefore
can contribute either allele to the offspring.
5. To determine the shape of the face, both people will flip
coins at the same time to determine
the genotype for the first trait. The coins should be flipped only
once for each trait. Each coin
flip represents the gamete contributed by the individual parent.
6. Continue to flip the coins for each trait listed in Data Table

1.
7. After each flip, record the trait of the offspring by placing a
mark in the appropriate box in the
table.
Note: Some information in the table has been simplified as some
of the traits are actually produced
by multiple genes. Trait 4 (hair type) exhibits incomplete
dominance.
8. Combine all of the characteristics into a drawing of the
offspring.
Exercise 2: Punnett Squares
The probability of genotype and phenotype expression in a new
generation of offspring can be
predicted mathematically. The use of Punnett squares,
developed by the geneticist Reginald
Punnett, is a method of organizing the mathematical
probabilities of genetic information that
will be passed down to subsequent generations. Punnett squares
are simple grids that show
the genotype of the mother (for one particular gene or trait) on
one axis of the gridline, and
the genotype of the father on the other axis of the gridline. Each
column or row of the grid
corresponds to one possible gene to be passed along to the
offspring.

The Punnett square for a simple monohybrid cross (examining
only one gene or trait of interest)
is shown in Figure 2. The father’s genes are shown on the top
row, and the mother’s genes are
shown in the left column. Note that the connecting cells for
each row and column in the grid
correspond to each possible gene combination from the father
and mother. In this example, each
parent is heterozygous (Aa) for the gene or trait.
Figure 2: Simple Punnett square for monohybrid cross
Father ( A a ) × Mother ( A a )
Father ♂
gametes A a
Mother
♀
A AA Aa
a Aa aa
When considering a dihybrid cross (examining two genes or
traits of interest at one time), all of
the options should be shown for the father and mother, as in the
monohybrid cross. See Figure
3. The father’s possible gene combinations are shown in the top
row, and the mother’s possible
gene combinations are shown in the left column. In this
example, each parent is heterozygous
(AaBb) for each of the two genes or traits.

Figure 3: Punnett square for a dihybrid cross
Father ( A a B b ) × Mother ( A a B b )
Father ♂
gametes AB Ab aB ab
Mother
♀
AB AABB AABb AaBB AaBb
Ab AABb AAbb AaBb Aabb
aB AaBB AaBb aaBB aaBb
ab AaBb Aabb aaBb aabb
If a trait is sex-linked, it is only present on the X or the Y
chromosome (recall the discussion
of X-linked traits earlier). If the trait is recessive, and is on the
X chromosome, then the trait is
only displayed in cases where there is only one X chromosome
(males) or where two of the X
chromosomes display the recessive trait (females). In Figure 4,
the father and the mother carry
the recessive trait. However, only the father is affected by the
recessive trait. When examining
the Punnett square, if the father and mother have a daughter,
there is a 50% possibility that she
will display the recessive trait. In addition, if they have a son,
there is also a 50% possibility that
he will display the recessive trait.
Figure 4: Punnett square for a sex-linked trait

Father ( Xg y ) × Mother ( XG Xg )
Father ♂
gametes Xg y
Mother
♀
XG XG Xg XG Y
Xg Xg Xg Xg Y
Exercise 3: Pedigree charts
Mendel’s laws of inheritance cannot be tested as easily with
humans as with pea plants. Scientists
can control the reproduction of “model organisms” such as pea
plants and fruit flies in order
to precisely analyze how their traits are inherited. Controlled
human reproduction (known as
eugenics) is both unethical and illegal. Therefore, when
studying lineages of families, using
pedigrees (family trees that show phenotypic patterns) is often
the only way that geneticists can
study patterns of inheritance in humans. Inferences can be made
from these lineages. Figure
5 shows an example of a pedigree chart. Note the patterns of the
chart. In Figure 5, the trait is
caused by an autosomal (non-sex-linked) dominant trait.
Figure 5: Pedigree chart

prOcEDurE
1. Review the pedigree charts in Figures 6 - 8.
2. Using the key provided, label each individual with the correct
genotype.
Note: It is helpful to use a blank or a dash to stand in for an
unknown allele (as in A- or A_).
The three pedigree charts will exhibit one of the following
inheritance patterns:
a. Autosomal Dominant: If at least one chromosome has the
allele, the individual will be
affected (that it, it will show the phenotype of interest). In a
pedigree, one or both parents
will be affected, and many of the offspring will also be affected.
Individuals with genotypes
AA or Aa will be affected; those will aa genotypes will be
unaffected.
b. Autosomal Recessive: In order for an individual to be
affected, both alleles must be
present in the recessive form. This trait does not occur as often
as a dominant trait. In a
pedigree, affected parents will often have unaffected offspring,
and vice-versa.
c. X-linked Recessive: The trait is present only on the X
chromosome. In males, only one

gene is needed for it to be expressed. In females, the trait must
be homozygous in order
to be expressed. Females who are heterozygous for this trait
will not show have affected
phenotypes, but they are carriers of the trait. In a pedigree, the
trait will be most commonly
seen in males. XAXA and XAXa individuals will be normal
females but XAXa individuals will
carry the trait. An XAY male will be unaffected, while an XaY
individual will be affected.
Figure 6: Pedigree chart 1

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