BIO100 Biology LAB 11

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Lab 11

Human Genetics

Introduction: Connecting Your Learning

Humans are 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	None
Resources
Required	Lesson 14 Lab 11
Assignments
Required Materials	From the Lab Kit
PTC paper

Focusing Your Learning

Lab Objectives

By the end of this lab, 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, one will notice that certain features vary from individual to individual. These heritable features are called characters and 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 genetics 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 blonde-hair 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, the possible alleles of one parent are placed on the x axis in the table and the possible alleles of the other parent are placed 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). The student would be 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):

B B

b

Bb

Bb

b

Bb

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 ½ = ½ 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 b

b

Bb

bb

b

Bb

bb

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, it is said to be codominant. An example of a gene that expresses codominance is blood types. The four blood types are: A, B, O and AB. Type A and B are both expressed in the AB blood type, making the traits codominant. Partial or incomplete dominance occurs 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 sex-linked 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, they likely have red-green colorblindness.

Click on image to enlarge.

Sex-linked characteristics are typically denoted with a subscript (for example, Xb would be 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 karyotype is 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 cell in metaphase where the chromosomes are separated is then photographed. 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. It can then be determined 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 can be determined. A normal male karyotype is illustrated below.

Click on image to enlarge.

Some common abnormal human karyotypes are:

XO– Turner's syndrome – short female, sexually immature. XO means there is only one X chromosome, andthe 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's (Mongolism) – large tongue, "slanted eyes, wide/webbed neck, mental retardation.Trisomy 21 means there are 3 chromosomes in the 21st position.

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. Characters 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 co-dominant. Absence of either A or B is called O. The Rh blood group of an individual is determined by the presence of certain proteins as part of the cell membrane of red blood cells. A person is Rh positive if they possess the D antigen and are termed Rh negative if they do not have a D antigen.

Widows Peak: The structure of the hairline on the forehead is genetically determined. The dominant gene W produces a growth of hair downward to a point above the nose and is called a widow's peak.

Ear Lobes: The lobe (bottom part) of the ear may be totally attached to the face, or it may hang 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, they are able to bend outward the terminal section of their thumb to almost a 45-degree angle. This is known as a hitchhiker's thumb. Dominant gene H individuals cannot bend their thumb.

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

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

phenylthiocarbide (PTC). This gives 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 called 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 that the student possesses.

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 the student's 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 student's 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 individual’s chromosomes presented by clicking on the link below. Be sure to record the case number provided as it will be needed later. Please click on the link at the top of the activity and print off the chromosomes. Make sure you print out the karyotype as soon as you open the lab. There are multiple karyotypes and will change each time you open the lab. 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. Subsequent pairs should be placed 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 will determine 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 ½ 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 on the link below to open up the online submit form for the laboratory. Paste the answer into the online dialog box by inserting the cursor in the submit box and simultaneously holding down the Ctrl and V keys. The answers should now appear in the box. 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 11

1. List whether the student's 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 blood type if the student is heterozygous or homozygous? Explain. (5 points)

3. Select a trait of interest.

A. What was the trait? (1 point)

B. What is the phenotype for the trait, and 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. (Note: you do not need to submit the punnett square itself, only the results of the punnett square).

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 color-blind. 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. From the Human Karyotype Activity, what was the Karyotype Group number? What was the result of the 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 would be 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 with a population of 120 individuals (n = 120): (4 points)

A. What were the possible genotypes of the offspring?

B. What is the probability of males having hemophilia out of the entire population?

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 student has purple and white pansies. The student 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 student decide 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 usefulto the student, or how can the student apply this knowledge to everyday life as a non-scientist? The application paragraph will be graded according to the rubric below. (20 points)

	EXCELLENT	VERY	GOOD	FAIR	NEEDS
	GOOD	IMPROVEMENT
CRITICAL
THINKING
AND
APPLICATION	18-20	16-17	14-15	12-13	Below 12
OF
INFORMATION
(20 points)