BIO 156 Lab 7 Assignment Questions

27 Jun BIO 156 Lab 7 Assignment Questions

Question
Lab 7
Human Genetics
Introduction: Connecting Your Learning
The human body is composed of millions of cells; each cell contains genetic information that
makes an individual unique. Traits are inherited from a person’s parents in the form of alleles.
This lab will illustrate the differences between the genes that are inherited versus the traits that
are expressed. The importance of karyotypes and how they are used to identify possible
chromosomal aberrations will also be explored.
Resources and Assignments
Multimedia Resources

None

Required Assignments

Lesson 9 Lab 7
From the Lab Kit

Required Materials

PTC paper

Focusing Your Learning
Lab Objectives
By the end of this lesson, you should be able to:
1. Distinguish between an individual’s phenotype and genotype.
2. Explain how traits are passed from parent to offspring.
3. Utilize Punnett squares to determine the genotypic and phenotypic ratios of specific
traits.
4. Distinguish between the terms codominance and incomplete dominance.
5. Explain why sex-linked diseases are more common in male humans.
6. Determine the genetic makeup of certain characteristics.
7. Interpret a karyotype to determine an individual’s sex and to identify possible
chromosomal disorders.
Background Information

When observing organisms, certain features vary from individual to individual. These heritable
features are called characters; an example is hair color. Characters have several variations,
which are called traits. Using the example of hair color as a character, traits for hair color would
be red, black, brown, or blonde. The trait that is expressed is called a person’s phenotype. For
example, someone with brown hair has the brown phenotype. Recall from the cell division
laboratory conducted earlier in the course that genes contain different versions, called alleles.
The allelic combination that comprises a character is called the genotype. For example, the
genotype may be BB (two brown alleles). When specifying the genotype for a specific character,
a capital letter is used to indicate a dominant trait (B), and a lowercase letter is used to indicate
a recessive trait (b).
The expression of these traits (or phenotypes) depends upon the interaction of the alleles. If a
trait is expressed and it is found on only one of the chromosomes of the pair, it is said to be
dominant over the other trait on the other chromosome of the homologous pair. For example,
brown hair (B) is dominant over blonde (b). If someone has the brown-hair trait and the blondehair trait, that person will have brown hair (Bb).
Traits are passed from parent to offspring; each parent donates one allele for a specific gene.
This is the result of sexual reproduction, where the gamete (sperm, egg, or pollen, for example)
contains only one copy of each chromosome pair (homolog). If both chromosomes originally
had the same characteristics (genes), then all gametes produced would have that characteristic.
For example, if the trait is homozygous dominant, all resulting gametes will have the same
allele. If the original alleles were different, however, then two different gametes could be
produced. For example, if the trait is heterozygous, one gamete could have the dominant allele
while the other could have the recessive allele.
If the composition of possible gametes is known, the frequency of a character in the offspring
can be calculated. By calculating the frequency of a gamete trait and then multiplying it by the
frequency of the trait in the other sex, the frequencies of the combinations in the offspring can
be determined. Application of these concepts was discussed in the genetics laboratory where
Punnett squares were employed to determine the frequencies of genotypic and phenotypic
ratios of specific characteristics in offspring. As a reminder, when creating a Punnett square,
place the possible alleles of one parent on the x axis in the table and place the possible alleles
of the other parent on the y axis in the table.
For example, using the hair color example provided earlier, if a mother has blonde hair and the
father has brown hair, the chances of their baby being born with blonde hair can be determined.
For the mother, since blonde hair is a recessive trait, she possesses the following genotype for
hair color – bb. In comparison, the father has brown hair, which is the dominant trait. Since it is
dominant, he can either be homozygous for brown hair (BB), or he can be heterozygous for
brown hair (Bb). To determine the frequency for the trait (blonde hair), there will be two
possibilities. For the first calculation, use the mother’s genotype of blonde hair (bb) and
calculate the frequency based on a homozygous dominant father (BB). For the mother, the
frequency of the blonde trait is one (because the only allele she possesses is the blonde trait).
For the father, the frequency of the blonde trait is zero (because there are no blonde alleles if
his genotype is homozygous dominant). This means multiplying 1 X 0, which is 0. The

calculation indicates that no children would be born with blonde hair. To confirm, the student can
create a Punnett square, as seen below.
Homozygous, dominant father (BB)

Genotype: 100% heterozygous dominant (Bb)
Phenotype: 0% chance of having a baby with blonde hair.
In the second scenario, calculate the possibility of having a child with blonde hair if the father is
heterozygous dominant (Bb). Using the same calculation above, the frequency of the blonde
trait from the mother’s genotype is one, and the frequency of the blonde trait from the father’s
genotype is one-half (because half of the alleles are for blonde hair). The calculation to
determine the frequency of blonde hair in their children is 1 x = or 50%. This means that, if
the father is heterozygous dominant, there is a 50% chance that their child could be born with
blonde hair. To confirm, you can create a Punnett square, as seen below.
Heterozygous, dominant father (Bb)
B
Genotypic ratio: 50% heterozygous dominant (Bb) and 50% homozygous recessive (bb)
Phenotypic ratio: 50% chance having a baby with blonde hair.
If both traits are expressed, this is called codominant. An example of a gene that expresses
codominance is blood types. The four blood types are A, B, O, and AB. Types A and B are both
expressed in the AB blood type, making the traits codominant. Partial or incomplete
dominanceoccurs if an intermediate trait is expressed. This is often seen in flower colors. Some
flowers are pink if they have one red allele and one white allele. In this case, the flower does not
completely express the red or white trait; instead, it expresses an intermediate trait of pink, as
seen in the image below.

Click on image to enlarge.

Sex-linked (X-linked) traits are associated with the X chromosome, which is the female sex
chromosome. Since the X chromosome contains many more genes than the Y chromosome,
most sex-linked genes are found on the X chromosome. Since there is no corresponding trait
found on the Y chromosome for a given characteristic, the recessive trait on the X chromosome
is expressed. With sex-linked traits, males frequently exhibit the traits, while expression in
females is rare. Many of the sex-linked traits are recessive. Example diseases caused by sexlinked traits include hemophilia and color-blindness. Red-green color-blindness is the most
common type of color blindness. The image below shows a simple test that can be used to
determine if an individual has red-green color-blindness. If an individual cannot see the green
numeral in the image, he likely has red-green colorblindness.

Click on image to enlarge.

Sex-linked characteristics are typically denoted with a subscript (for example, Xb is used to
indicate an X chromosome that contains the trait for color-blindness). Females who posses one
copy of the recessive trait will not express the characteristic because they are heterozygous and
possess one chromosome with the normal trait. As a result, these individuals do not express the
characteristic associated with the sex-linked traits and they are called carriers. In order for a
female to express the trait, she would need to possess two copies of the trait: one on each
chromosome.

While understanding genetic makeup is important, it is also important to understand how
chromosome numbers affect individuals. A karyotypeis a pictorial representation of the
chromosomes of an individual. A blood sample is taken and the cells are stimulated to undergo
mitosis. A smear is made, stained, and observed under the microscope. A photograph is taken
of a cell in metaphase where the chromosomes are separated. The image is enlarged and then
the chromosomes are matched up. The chromosomes are arranged based on size, from largest
(Pair 1) to smallest (Pair 22), followed by the sex chromosomes. Scientists can then determine
what sex the individual is, if the normal number of chromosomes is present, and if the
chromosomes are of normal length (no deleted portions or added/rearranged portions). Through
this process, some genetic disorders are determined. The image below illustrates a normal male
karyotype.

Click on image to enlarge.

Some common abnormal human karyotypes are the following:
XO Turner’s syndrome short female, sexually immature. XO means there is only one X
chromosome, and the other is absent.
XXY Klinefelter’s syndrome sterile male, poorly developed sexual structures, enlarged
breasts
XXX Trisomy X normal to resembling Turner’s, mental retardation
Trisomy 21 Down Syndrome (Mongolism) large tongue, slanted eyes, wide/webbed neck,
mental retardation
XYY Supermale tall, aggressive, high frequency of mental institutionalization
As part of the study on the Animal Kingdom, some common characteristics of humans will be
examined to try to predict the genotype for each characteristic. Characteristics that will be
examined include:

Blood Type: The ABO blood type of an individual is determined by the presence of
certain glycoproteins on the surface of the red blood cells. The A and B antigens
(characteristics) are codominant. Absence of either A or B is called O. The presence of
certain proteins as part of the cell membrane of red blood cells determines the Rh blood
group of an individual. A person is Rh positive if he/she possesses the D antigen and is
termed Rh negative if he/she does not have a D antigen.

Widow’s Peak: Genetics determine the structure of the hairline on the forehead. The
dominant gene W produces a growth of hair downward to a point above the nose called
a widow’s peak.

Ear Lobes: The lobe (bottom part) of the ear is either totally attached to the face, or it
hangs freely. The free hanging ear lobe is due to a gene F, which is dominant.

Tongue Rolling: The ability to roll the edges of one’s tongue up when the tongue is
protruded (to appear as a U when viewed from the front) is due to a dominant gene R.

Hitchhikers Thumb: If an individual is recessive for the h gene, he or she is able to bend
outward the terminal section of the thumb to almost a 45 angle. This is known as a
hitchhiker’s thumb. Dominant gene H individuals cannot bend their thumbs.

Thumb Crossing: If one clasps their hands together naturally without any thought, the
thumb that most frequently is on top is genetically inherited. Individuals who exhibit the
left thumb over the right have the dominant gene F. (Some references call this gene C.)

Little Finger Bend: The B gene is dominant over b and is evidenced by the little finger
terminal portion being bent/curved towards the fourth finger.

PTC Taster: An individual can inherit the ability to taste certain chemicals. One such
compound is phenylthiocarbamide (PTC). It has a bitter taste. Individuals are considered
homozygous for T if they can taste the compound quickly; heterozygous for T if they can
taste the compound after a short while; and homozygous for t if they cannot taste it at all.
Those who are TT are called tasters while those who are tt are non-tasters.

Mid-digit hair: Some people have hair on the second (middle) digit of their fingers. The
presence of hair on the middle digit is a dominant trait (H) and the absence of hair is a
recessive trait (h).

Facial dimples: One or more dimples on the face when a person smiles is a dominant
trait (D) while no dimples on the sides of the face when a person smiles is a recessive
trait (d).

Freckles: The presence of freckles (F) is dominant over the absence of freckles (f).

Chin cleft: The presence of a dimple in the chin, called a chin cleft, is dominant (C) over
the absence of a cleft (c).

Additionally, a karyotype for an unknown individual will be constructed and any possible genetic
abnormalities will be identified.
Procedures
1. Determine the various characters expressed in your phenotype.

A. Record whether the student is positive or negative for the following traits. (Note:
For blood type, record the student’s blood type instead of whether the student
possesses the trait.) These results are the phenotypes.
Blood type: A, B, or O (Record blood type.)
Widow’s peak
Free ear lobes
Tongue rolling
Hitchhiker’s thumb
Left thumb dominance
Little finger bend
PTC taster
Mid-digit hair
Facial dimples
Freckles
Cleft chin
B. List the possible genotype/s for each trait from Part A.
C. Select three traits to investigate further. Record whether the parents also
possess the traits or not. Then, create Punnett squares to see what the possible
genotypes and phenotypes would be for the trait in children born to the student’s
parents. For example, if the student has a widow’s peak, it is a dominant trait.
The student’s Mom also has a widow’s peak, but the student’s Dad does not.
Knowing that the absence of a widow’s peak is recessive, the student’s Dad’s
genotype must be ww. Since a widow’s peak is a dominant trait, the student’s
Mom’s genotypes could be Ww or WW. Therefore to determine possible
presence of a widow’s peak in the children, the student will create two Punnett
squares – one with a cross of ww x WW and a second with a cross of ww and
Ww. Record the results in the form of a ratio or percent chance of the trait being
present in children with each cross.
D. Create a karyotype of an unknown individual’s chromosomes presented by
clicking on the activity below. Be sure to record the case number provided as it
will be needed later. If the image does not load properly or the chromosomes are
difficult to match due to the size, please click on the link at the top of the activity
and print off the chromosomes. To construct the karyotype, observe the
chromosomes for similarities and match them by pairs of two. Drag the first two

chromosomes that match into the first window. When constructing karyotypes,
the homologous chromosomes should begin with the largest chromosome pair in
the first window. Place subsequent pairs in the windows according to the next
largest size. There should be a total of 22 pairs of matching chromosomes; the
twenty-third pair of chromosomes determines the sex of the individual. If using
printed chromosomes, each chromosome should be cut out and then matched
with the corresponding chromosome. The chromosome pairs should be taped or
glued to a clean sheet of 8 x 11 inch printer paper, with pairs arranged from
largest to smallest. Once the karyotype is constructed, identify the sex of the
individual, as well as the chromosomal disorder they have. Note: Not all
individuals will have a chromosomal disorder.
E. Create a Punnett square that determines the probability of male offspring having
hemophilia when the father has hemophilia (Y Xh) and the mother is a carrier
(XH Xh). Record the results.

Assessing Your Learning
Compose answers to the questions below in Microsoft Word and save the file as a backup copy
in the event that a technical problem is encountered while attempting to submit the assignment.
Make sure to run a spell check. Copy each answer from Microsoft Word by simultaneously
holding down the Ctrl and A keys to select the text, and then simultaneously holding down the
Ctrl and C keys to copy it. Then, click the link on the Lab Preview Page to open up the online
submit form for the laboratory. Paste the answer for the first question into the online dialog box
by inserting the cursor in the box and simultaneously holding down the Ctrl and V keys. The
answer should now appear in the box. Repeat the process for each question. Review all work to
make sure that all of the questions have been completely answered and then click on the
Submit button at the bottom of the page.
LAB 7
1. List whether the student was positive or negative for each characteristic and include
whether the characteristic is dominant or recessive. (6 points)
a. Blood type
b. Widow’s peak
c. Free ear lobes
d. Tongue rolling
e. Hitchhiker’s thumb

f.

Left thumb dominance

g. Little finger bend
h. PTC taster
i.

Mid-digit hair

j.

Facial dimples

k. Freckles
l.

Cleft chin

2. Can the student tell from the blood type if the student is heterozygous or homozygous?
Explain. (5 points)
3. Select a trait of interest.
a. What is the trait? (1 point)
b. What is the phenotype for the trait? Is this the dominant or recessive allele for the
trait? (2 points)
c. What are the possible genotypes for the parents? (2 points)
d. Include the results of one Punnett square, showing a possible combination
between alleles for the trait from the parents by filling in the genotypes according
to the numbers in the square, below.
1

2

3

5

6

4

7

8

e. Based on the cross, what percent of children born to the parents would express
the trait? (1 point)
4. Create a Punnett square to determine the possibility of a couple having a color-blind
child if the mother has the recessive trait on one X and the father is colorblind. HINT: Use Xb to indicate an X with the color-blindness trait. How many female
offspring will be color-blind? How many male offspring? (5 points)

5. What was the group number of the student’s karyotype? What was the result of the
student’s karyotype? (Include the sex and the chromosomal disorder, if applicable. If
there was no chromosomal disorder, the student must state that the individual was
normal.) (5 points)
6. Give an example of a situation in which it is important to create a karyotype for an
individual. Explain. (5 points)
7. Genetically speaking, why is it important not to mate with a close relative? Explain. (5
points)
8. Does a karyotype tell all of a person’s genetic characteristics? Explain. (5 points)
9. Why is a photograph of cells in metaphase utilized when constructing a karyotype? (5
points)
10. What does it mean to be a carrier of a genetic defective characteristic? When might it be
important to know if one is a carrier? (5 points)
11. From the hemophilia procedure: (4 points)
a. What were the possible genotypes of the offspring?
b. What is the probability of males having hemophilia?
c. How many females would have hemophilia?
d. How many carriers would there be?
12. Explain why more males tend to suffer from X-linked disorders than females. (5 points)
13. The student has a friend that knows the student is taking biology, and she is confused
about her blood type. Her blood type is O, but her dad is A and her mother is B. She
asks the student if it is possible for her parents to have a child that is O. Explain the
answer to her. (5 points)
14. In a flower garden, the gardener has purple and white pansies. He notices that a new
pansy has sprouted. When it finally flowers, the pansy is lavender. Explain how this
happened. (5 points)
15. With a botanist friend’s help, the gardener decides to cross the lavender pansy with the
white pansy. Will this result in any purple pansies? Explain. (5 points)
16. (Application) How might the information gained from this lab pertaining to human
genetics be useful to a student enrolled in a healthcare related profession? (20 points)
Have You Met The Objectives For This Lesson?