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Analysis of Human Traits
HASPI Medical Biology Lab 22b
Background/Introduction
Using Statistics and Probability to Analyze Genetic Traits
The occurrence of natural selection is necessary in order for evolution to occur. Natural
selection requires variation in genetic information as well as variation in how that information
is expressed. For example, if an isolated human population had only one trait--blonde hair-every individual in that population will have blonde hair. Because there is no variation in
genetic information or expression, natural selection cannot occur. If a mutation occurred in
a single individual to produce brown hair, there would now be variation in the genetic
information and expression: blonde and brown hair. If the environment were more
conducive to brown-haired individuals, then brown hair would become more common.
Traits that positively affect the survival of a population are more likely to be reproduced.
How can we quantitatively measure whether natural selection is occurring within a
population? Statistics and probability can be used to support explanations that organisms
with advantageous heritable traits tend to increase in proportion, as opposed to organisms
that do not have the advantageous trait. There are several statistical and probability-based
models that can be used to analyze and predict trait expression:
Punnett Squares
A Punnett square is a chart used to predict the genotype (genetic characteristics) and the
phenotype (physical characteristics) of the children in a cross between two parents. For our
purposes, Punnett squares are capable of predicting the probability of offspring inheriting
genetic traits, or conditions, from their parents. Genetic variations that cause these
conditions can be found on autosomes (chromosomes that are not sex-linked) or on the sex
chromosomes. Traits found on the autosomes can be dominant, recessive, or demonstrate
more complex interactions. Traits found on the sex chromosomes are called sex-linked. The
following examples demonstrate how to use a Punnett square to determine genotype and
phenotype for autosomal, sex-linked, and two-factor crosses.
How to Perform a Punnett Square
Steps
Step 1. Designate a letter to represent the
condition. A capital letter represents the
dominant trait, and a lowercase letter represents
the recessive trait. If the traits are sex-linked, the
capital or lowercase letter is represented as a
subscript next to either the X or Y sex
chromosome.
Step 2. Determine the genotypes of each parent.
If a parent is homozygous dominant, he/she has
two (same) dominant traits (DD); if the parent is
homozygous recessive, he/she has two (same)
recessive traits (dd); and if the parent is
heterozygous (different), he/she has both a
Example
Autosome Trait
D = dominant
d = recessive
Sex-linked Trait
XD = dominant
Xd = recessive
Autosome Trait
mom is heterozygous and dad is homozygous recessive
Dd x dd
Sex-linked Trait
mom is heterozygous and dad is recessive
XDXd x XdY
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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dominant and a recessive trait (Dd).
Step 3. Draw the Punnett square. If we
are only predicting the probability of one
trait, it is a monohybrid cross. If we are
predicting the probability of two traits, it
is a dihybrid cross.
Step 4. Write the possible traits that each
parent could pass to the offspring along
the sides and top of the Punnett square.
Step 5. Fill each box of the Punnett
square by recording the letters from
above and to the left of each empty
box. Uppercase letters are always
placed before lowercase letters.
Monohybrid Cross
Autosomal Trait
Dd x dd
Sex-linked Trait
XDXd x XdY
d
d
D Dd Dd
d dd dd
XD
Xd
2/4 offspring Dd = 50%
2/4 offspring dd = 50%
Xd
Y
XDXd XDY
XdXd XdY
¼ offspring XDXd = 25%
¼ offspring XdXd = 25%
¼ offspring XDY = 25%
¼ offspring XdY = 25%
Dihybrid Cross
Autosomal Traits
DdHh x DdHh
Step 6. Determine the probability of the
offspring receiving a trait. List the
possible genotypes of the offspring
below the Punnett square. The genotype
results can be written as fractions or
percentages. Once the genotype is
known, the phenotype of offspring can
also be determined.
DH
Dh
dH
dh
DH
DDHH
DDHh
DdHH
DdHh
Dh
dH
dh
DDHh DdHH DdHh
DDhh DdHh Ddhh
DdHh ddHH ddHh
Ddhh ddHh ddhh
1/16 DDHH = 6.25%
1/16 DDhh = 6.25%
1/16 ddHH = 6.25%
1/16 ddhh = 6.25%
2/16 DDHh = 12.5%
2/16 DdHH = 12.5%
2/16 Ddhh = 12.5%
2/16 ddHh = 12.5%
4/16 DdHh = 25%
Pedigrees
A pedigree is a diagram that uses symbols to represent family and genetic relationships.
Pedigrees make it easier to visualize inheritance patterns within a family, and can be used to
determine how a trait, or condition, is inherited. The following symbols and lines are used to
represent individuals and relationships within a pedigree.
Common Pedigree Symbols
570
= normal male
= normal female
= parents
= afflicted male
= afflicted female
= generation
= carrier male
= carrier female
= siblings
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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= deceased before
reproducing male
Date:
= deceased before
reproducing
= adoption
female
A pedigree is particularly useful when determining how genetic diseases are inherited within
a family. Individuals who know their chances of having children that may inherit a difficult
disease have the chance to make more informed decisions in their choices to have children.
The following is an example of a simple pedigree with a mother who is afflicted with a
genetic disease, and a father who is normal over three generations.
Example Pedigree
Generation I
Generation II
Generation III
Hardy-Weinberg Equilibrium
The Hardy-Weinberg formula is used to analyze traits in populations, and detect whether a
change has occurred in the frequency of those traits. In this way, the Hardy-Weinberg
formula is capable of determining whether evolution has occurred. A change in the gene
frequencies of a population over time can be used to determine whether the traits in a
population are changing. If there is little or no change, the population is in equilibrium and
no evolution is occurring.
The Hardy-Weinberg Formula(s)
p+q=1
p2 + 2pq + q2 = 1
p = frequency of dominant allele
q = frequency of recessive allele
p2 = percentage of homozygous dominant
q2 = percentage of homozygous recessive
2pq = percentage of heterozygous
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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Hardy-Weinberg Example Problems
Example Problem 1
16% of the Alaskan population has a recessive
condition that makes the individuals sensitive to
sunlight. Determine how many individuals in the
population are heterozygous for this condition.
1. Calculate the frequency of a single
recessive allele (q)
If q2 = 16% = 0.16, then q = 0.4 (square root of
0.16)
The frequency of the recessive allele as a percentage is
40% (0.4 x 100 = 40%)
2. Calculate the frequency of a single
dominant allele (p)
Since q = 0.4, use the p + q = 1 equation
p + 0.4 = 1  p = 0.6
The frequency of the dominant allele as percentage is 60%
3. Calculate the frequency of the remaining
genotypes (homozygous dominant and
heterozygous)
homozygous dominant
if p = 0.6 then p2 = 0.6 x 0.6 = 0.36
heterozygous
if p = 0.6 and q = 0.4, then
2pq = 2(0.6)(0.4) = 0.48
The homozygous dominant frequency as a percentage is:
_____% ; Heterozygous frequency as a percentage is:
_____%
Check your answers by adding your p2, 2pq, and q2
values. (Remember: p2 + 2pq + q2 = 1)
Your values should equal 1.
Example Problem 2
A condition causing extremely brittle nails is
recessive. In a village population, it is observed that
654 people have normal nails, and 86 people have
brittle nails. How many people are homozygous
dominant, heterozygous, and homozygous
recessive?
1. Calculate the frequency of a single
recessive allele (q)
There are 86 people with the condition,
which can be used to determine q2:
86/740 total = 0.12 (or 12%) = q2
2
if q = 0.12, then q = 0.35 (square root of
0.12)
2. Calculate the frequency of a single
dominant allele (p)
Since q = 0.35, use the p + q = 1 equation
p + 0.35 = 1  p = 0.65
3. Calculate the frequency of the remaining
genotypes (homozygous dominant and
heterozygous)
homozygous dominant
if p = 0.65 then p2 = 0.65 x 0.65 = 0.42
42% of the village population = 0.42 x 740 = 310 individuals
heterozygous
if p = 0.65 and q = 0.35, then
2pq = 2(0.65)(0.35) = 0.46
46% of the village population = 0.46 x 740 = 340 individuals
Number of homozygous recessive individuals is: _____
The Chi-Square Test
While predicting the outcome of genetic crosses through probability is useful, it is only
meaningful to science when the data is compared to actual test results. If the test results do
not match the probability exactly, how can we determine whether the data collected from
an experiment is correct? We use a “goodness of fit” test, called the chi-square test, which is
a mathematical way to determine if the collected results are close enough to the expected
results to be significant (low probability of being wrong/by chance). The chi-square formula
is:
2
(
obs

exp)
2  
exp
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Analysis of Human Traits, HASPI Medical Biology Lab 22b
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FORMULA KEY
X = is the Greek letter chi Σ = sum
obs = observed results
exp = expected results
Use the following steps to conduct a chi-square test on expected and actual results of a
genetic cross. An example of each step has been provided for two parents who are
heterozygous for brown eyes (Bb x Bb).
1. Use a Punnett square to determine the expected phenotype results.
Phenotype
3 brown eyes (3/4 = 0.75)
1 blue eyes (1/4 = 0.25)
B
b
B
BB
Bb
b
Bb
bb
2. Add up the total number of observed results from the collected data.
Collected Data
Brown Eyes Blue Eyes Total Results
155
61
216
3. Now determine what the expected phenotype results would be with the total number
of observed results calculated.
Brown Eyes
Blue Eyes
Expected Results
0.75
0.25
Total Results
Expect Phenotype Results
216
0.75 x 216 = 162
0.25 x 216 = 54
4. Compare the expected and observed results (collected from experiment) using the
chi-square test.
Brown Eyes
Blue Eyes
X2 = (obs - exp)2 / exp
X2 = (155 - 162)2 / 162
X2 = (61 - 54)2 / 54
X2 = (-7)2 / 162
X2 = (7)2 / 54
X2 = 49 / 162
X2 = 49 / 54
X2 = 0.30
X2 = 0.91
X2 = 0.30 + 0.91 = 1.21
5. Compare the calculated chi-square value to Table 1 below. The degrees of freedom
(df) is simply the number of phenotype options minus 1. In a monohybrid cross, the df = 1
and in a dihybrid cross, the df = 3. A P-value of 0.05 means that the observed and
expected results did not deviate from each other more than 5%.
Table 1. Significance of
Results
Chi-Square Values
X2 = 1.21
2
Degrees of
X at a PThere are only 2 phenotypes so the df = 1,
Freedom (df) value of 0.05
which makes the significant chi value 3.84.
1
3.84
Since the chi value you calculated (1.21) is less than 3.84, the
2
5.99
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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4
5
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7.81
9.49
11.07
Date:
expected ratio of a genetic cross is correct. If the chi value you
calculated happened to be greater than 3.84, then the expected
ratio would be incorrect and there are likely additional factors
going on to create genetic variation for this gene.
HASPI Medical Biology Lab 22b
Purpose
When focusing on genetics to understand evolution, probability and concepts of statistics
are used to support explanations that organisms with an advantageous heritable trait tend
to increase, in proportion to organisms lacking this trait. Human genetic diseases caused by
mutations can appear to go against evolutionary theory, but not all mutations are
completely detrimental. Some traits produced by these mutations could even be
considered advantageous in certain environments. In this activity, you will have the
opportunity to use common statistic analysis models--Punnett squares, pedigrees, the HardyWeinberg formula, and the chi-square test--to analyze human traits and hypothesize whether
some of the traits could relay an advantage.
Materials
Calculator
Directions
For each statistical analysis, a description and directions are provided followed by practice
problems with real human mutations.
1. What’s That Smell?
Trimethylaminuria is an autosomal recessive condition that causes a pungent smell. Individuals
with this condition cannot break down a compound known as trimethylamine, whose smell has
been described as rotting eggs, rotten fish, garbage, or even urine. Trimethylamine builds up in
urine, breath, and sweat creating a strong smell. Approximately 200,000 people in the United
States have trimethylaminuria.
http://www.lineaysalud.com/images/stories/sindromedeolorapescado(1).jpg
1a. Use a Punnett square to determine whether two
individuals who are heterozygous for the
trimethylaminuria trait could have a child with
trimethylaminuria.
574
1b. Use a Punnett square to determine the chances
of a man with trimethylaminuria and a woman who is
normal (TT) having a child with trimethylaminuria?
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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1c. Could this trait possibly
be advantageous in a
specific environment?
Explain your answer.
2. Origins of the Werewolf Myth
Hypertrichosis, also known as “werewolf syndrome,” is a genetic condition that causes
excessive body hair growth. Those who have this condition exhibit hair on the face,
shoulders, and ears. There are two types of hypertrichosis – the most severe form is
hypertrichosis lanuginose which is an autosomal dominant condition, while generalized
hypertrichosis is less severe and sex-linked.
http://cdn2.top10hut.com/wp-content/uploads/2013/02/1326947669_03657000.jpg
2a. A mother is heterozygous for hypertrichosis
lanuginose. What are the chances of her having a
child with hypertrichosis lanuginose if the father does
not have the condition? Use a Punnett square to
explain your answer.
2b. A man has generalized hypertrichosis and has
married a woman who does not carry the trait. He is
worried about his children having the trait. What are
the chances of them having children with
generalized hypertrichosis?
2c. Could this trait possibly
be advantageous in a
specific environment?
Explain your answer.
3. Blood Type and Heart Disease Risks
Recent research has proposed a link between heart disease and blood types.
Individuals with Type O blood have the lowest risk, Type A had a 6% increase in
risk, Type B had a 15% increase in risk, and Type AB had a 23% increase in risk.
There are actually 3 traits, or alleles, that contribute to blood type – A (iA), B (iB),
and O (i). Since there are more than two alleles, blood types are considered to
have multiple alleles. We can only inherit two alleles, and how these alleles
combine determine our blood type. Both the A allele and the B allele are
dominant, and they both show up in the phenotype. This is called codominance.
http://www.medicalook.com/diseases_images/coronary_heart_disease.jpg
3a. A couple both have Type AB blood. They are
worried that all of their children will also be Type AB
and have a high risk for heart disease. What
3b. A child has Type O blood. What blood types
could the child’s parents have?
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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percentage of their children could have Type AB?
3c. Could this trait possibly
be advantageous in a
specific environment?
Explain your answer.
4. A Real Life Genetic Superhero?
Two relatively new and rare mutations could be capable of making a genetic
superhero. The first mutation is autosomal dominant and impacts the LRP5 protein,
which regulates bone mineral density. The mutation causes the protein to increase
the density of bones making them incredibly strong. The second mutation is
autosomal recessive and impacts the myostatin protein, which limits muscle growth.
Individuals with this mutation are capable of drastically increasing muscle growth
and strength. The combination of incredibly strong bones and strength in a single
individual could be amazing!
http://1.bp.blogspot.com/_Dg1MLpkMYL0/TSuWc2fmLZI/AAAAAAAAAGA/RyZ7tih38vM/s1600/blog68.jpg
4a. Complete a dihybrid Punnett square to
determine if any of the children from the following
couple could create an individual with increased
bone mineral density and increased muscle
growth/strength.
Increased bone density = B
Normal bone density = b
Normal muscle growth/strength = M
Increased muscle growth/strength = m
4b. Could a father with increased bone density and
normal muscle growth, and a mother with increased
bone density and normal muscle growth, produce a
child with increased bone density and increased
muscle growth? Explain your answer using a Punnett
square.
The mother’s genotype is BbMm,
and the father’s genotype is bbMm.
576
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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4c. Could these traits
possibly be advantageous
in a specific environment?
Explain your answer.
5. Apolipoprotein Al-Milano
Apolipoprotein Al-Milano is a mutation in a protein that carries cholesterol within the body.
Individuals who have this mutation have less cholesterol in the bloodstream and have
a significantly reduced risk of coronary heart disease, which is caused by plaque
buildup in the arteries (atherosclerosis). In fact, to date none of the individuals with this
mutation have exhibited any signs or symptoms of coronary heart disease. While still rare,
3.5% of Limone sul Garda, a small village in Italy, carry this mutation.
http://www.thewellingtoncardiacservices.com/images/Heart/Coronorary-ArteryDisease2.jpg
The following pedigree displays the inheritance of the Apolipoprotein Al-Milano mutation in a family from
Limone sul Garda.
Generation I
Generation II
Generation III
2
1
4
3
10
11
5
12
6
13
8
7
14
15
9
16
17
5a. From looking at the pedigree, is the Apolipoprotein Al-Milano mutation autosomal dominant, autosomal
recessive, or sex-linked? Explain your answer.
5b. What are the genotypes of individuals 1 and 2? Explain your answer.
5c. How many of the children between individuals 1 and 2 should carry the Apolipoprotein Al-Milano
mutation? How many actually do carry the mutation according to the pedigree?
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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5d. How many of the children between individuals 3 and 4 should carry the Apolipoprotein Al-Milano
mutation? How many actually do carry the mutation according to the pedigree?
5e. How many of the children between individuals 8 and 9 should carry the Apolipoprotein Al-Milano
mutation? How many actually do carry the mutation according to the pedigree?
5f. Explain why the number of children predicted to carry the Apolipoprotein Al-Milano mutation is different
than the actual number of children produced by this family.
5g. Could this trait
possibly be advantageous
in a specific environment?
Explain your answer.
6. Hemophilia and The Royals
Hemophilia is a sex-linked condition in which
individuals are missing
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and platelets are needed in order to stop
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6b. Did any males in Queen Victoria’s line inherit hemophilia? If yes, who?
6c. Why is there a difference in the number of females/males that inherited this trait?
6d. If Maurice or Leopold were to marry a normal female (no hemophilia), what are the chances their
children would inherit hemophilia? What are the chances their grandchildren would inherit hemophilia?
6e. Could this trait
possibly be
advantageous in a
specific environment?
Explain your answer.
7. The Blue People
Methemoglobinemia is an autosomal recessive condition that causes a deficiency in
the protein that allows red blood cells to use glucose. The function of red blood cells is
to carry oxygen, and when they are carrying oxygen they are red in color. The
deficiency of the protein causes the red blood cells to carry oxygen differently and turn
blue in color. This creates a blue skin color, which is often worse under stressful
conditions. Methemoglobinemia has been found in Native Americans, Eskimos, and a
small population in Kentucky.
http://trialx.com/g/Congenital_Methemoglobinemia-4.jpg
7a. Using the following information, create a pedigree for the family to determine how methemoglobinemia
was inherited through a family. Carrie and her husband Mark are both healthy, but they have just had a son,
Jathan, born with methemoglobinemia. Neither Carrie nor Mark had ever heard of the disease. Carrie
knows very little about her mother’s extended family in Kentucky. Carrie’s mother, Sandra, passed away
when she was young, and Carrie moved with her father, Michael, to California. She has decided to visit her
Kentucky relatives to understand how her son may have inherited methemoglobinemia. Visiting with her
great grandmother, Vivian, gave her the following information:
Vivian’s mother and father, Mary and Luke, looked normal and never turned blue. Vivian and her two sisters,
Anna and Jemma, also looked normal, but their brother Jeb always turned slightly blue when he was
worried. Jeb and Anna never had any children. Jemma married Tom from down the road and they had five
children: two boys and three girls. All of Jemma and Tom’s children looked normal. Vivian married Sam from
across the river. They had two children, Robert and Sandra, who were both normal. Robert died when he
was 8 from pneumonia. Sandra had one child, Carrie.
Using this information, create a pedigree of Carrie’s family and relatives to determine how her son ended up
with methemoglobinemia.
Analysis of Human Traits, HASPI Medical Biology Lab 22b
579
Name(s):
Period:
Date:
7b. Could this trait
possibly be advantageous
in a specific environment?
Explain your answer.
8. Backwards or Forwards Evolution: Uner Tan Syndrome
Uner Tan Syndrome is an extremely rare autosomal recessive condition in
which individuals travel on all fours, speak primitively, and have varying
levels of mental retardation. It was only recently discovered and is
localized in western Asia. While there have been multiple accounts of
Uner Tan Syndrome, currently only a single family in Turkey is exhibiting
symptoms of this genetic condition.
http://blogs.plos.org/neuroanthropology/files/2010/09/turkey-walking-on-all-fours-1.jpg
8a. A hidden population has been discovered in Turkey. Researchers have calculated that 22% of the
population have Uner Tan Syndrome. Using the Hardy-Weinberg formulas, calculate what percentage of the
population are heterozygous carriers for this condition. Show your work.
8b. Could this trait
possibly be advantageous
in a specific environment?
Explain your answer.
9. Human Claws: Ectrodactyly
Ectrodactyly is an autosomal dominant disorder that causes a cleft, or
split, where the middle finger or middle toe should form. This creates a
limb, or limbs, that appear to form a claw, which is why this condition
is also known as “lobster claw syndrome.” Some individuals also suffer
from hearing loss. Approximately 1 in 90,000 births results in a child
with this condition.
http://withfriendship.com/images/i/44490/with-ectrodactyly-usually.jpg
Analysis of Human Traits, HASPI Medical Biology Lab 22b
580
Name(s):
Period:
Date:
9a. The town of Freunberg has a population of 3,523. Approximately 652 individuals in this population have
extreme ectrodactyly, which is a homozygous dominant condition. Using the Hardy-Weinberg formulas,
calculate what percentages of the population are homozygous dominant, heterozygous, and homozygous
recessive. Show your work.
9b. Could this trait
possibly be advantageous
in a specific environment?
Explain your answer.
10. Sickle Cell and Malaria Resistance
Sickle cell anemia is an autosomal recessive condition that causes red blood cells to
Sickle Red Blood Cell
collapse and appear curved in shape. This shape limits the ability of red blood cells
to carry oxygen, and individuals with sickle cell anemia experience varying degrees
of pain, shortness of breath, and blood flow blockage. Sickle cell is an interesting
disease because, while it is recessive, individuals who are heterozygous experience
mild symptoms of sickle cell anemia. This is called incomplete dominance and occurs
because half of the red blood cells are created with the normal allele, and are normal, Plasmodium Parasite
while the other half of the red blood cells are created with the sickle cell allele, and
are sickle-shaped. Even more interesting is the fact that individuals who are
heterozygous for sickle cell anemia are protected from malaria, a serious infectious
disease passed by mosquitoes and caused by the Plasmodium parasite. Malaria
infections are very common in areas of Africa, and from recent research it is estimated
that 10-40% of the populations in these areas carry the sickle cell allele.
https://www.sciencenews.org/sites/default/files/14307 ; http://compulenta.computerra.ru/upload/iblock/88b/o_plasmodium%20falciparum.jpg
10a. In Kenya, approximately 16% of the population
has sickle cell anemia (homozygous recessive). What
percentage of the population will have resistance to
malaria from the sickle cell allele? Show your work.
10b. A genetic research project in Ethiopia
determined that 68% of the population carries the
normal allele for red blood cells. What percentage of
the population carries the sickle cell allele? Show your
work.
10c. What percentages of the population are
homozygous dominant, heterozygous, and
homozygous recessive?
Analysis of Human Traits, HASPI Medical Biology Lab 22b
581
Name(s):
Period:
Date:
10d. Could this trait
possibly be advantageous
in a specific environment?
Explain your answer.
11. The Dominance of Marfan Syndrome
Marfan Syndrome is an autosomal dominant condition that affects connective tissues within
the body. Due to the fact that connective tissues are present throughout the body, the
mutation causes a large variety of signs and symptoms. A few common signs include
disproportionate limbs, tall and slender build, protruding breastbone, crowded teeth, scoliosis,
heart murmurs, and nearsightedness. Approximately 1 in 10,000 births result in a child with
Marfan Syndrome.
http://i.imgur.com/3fXmSxt.jpg
11a. In a research project, children of individuals that were known to be heterozygous for Marfan Syndrome
were tested for the disease. The following observed results were collected:
Marfan Syndrome (Mm)
378
Normal (mm)
412
Perform a chi-square test to determine whether these results support that Marfan Syndrome is a dominant
disease. Show your work.
11b. Could this trait
possibly be advantageous
in a specific environment?
Explain your answer.
12. Is Cystic Fibrosis Really Recessive?
Cystic fibrosis is an autosomal recessive disease caused by a mutation in a common
cell membrane protein. This mutation creates a problem with how substances move
into and out of cells, causing a buildup of substances such as sweat and mucus in
body tissues and organs. A common symptom of cystic fibrosis is an overabundance
of mucus in the lungs. The life expectancy for individuals with cystic fibrosis was
historically very low (early 20s), but has improved due to medical discoveries.
Analysis of Human Traits, HASPI Medical Biology Lab 22b
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Name(s):
Period:
Date:
Approximately 1 in 3,000 births results in a child with cystic fibrosis.
http://medicalstuttering.com/wp-content/uploads/2013/05/Cystic-fibrosis.jpg
12a. In a 2008 study, there were 30,000 people in the United States with cystic fibrosis. The approximate total
population of the United States at the time was 305,000,000. Perform a chi-square test to determine whether
these results support that cystic fibrosis is a recessive disease. Show your work.
12b. Could this trait
possibly be advantageous
in a specific environment?
Explain your answer.
13. A New Dwarfism
Dwarfism is a condition characterized by individuals who are short in stature caused by
a mutation that slows or delays growth. Technically, dwarfism is defined as a height of
less than 147 cm. There are actually different dwarfism mutations. Achondroplasia is
an autosomal dominant condition, while diastrophic dysplasia (DTD) is an autosomal
recessive condition. Approximately 70% of all dwarfism is caused by achondroplasia.
http://geneticmutationruebe3.wikispaces.com/file/view/little-people-big-world31.jpg/309905494/little-people-big-world31.jpg
13a. A new form of dwarfism has been discovered in a population in a remote area of Australia. Out of a
population of 980, researchers counted 72 individuals with dwarfism. Using these values and the chi-square
test, determine whether the condition is dominant or recessive. (HINT: You will need to compare your results to
phenotype ratios for a dominant condition AND a recessive condition.) Show your work and explain your
answer.
Analysis of Human Traits, HASPI Medical Biology Lab 22b
583
Name(s):
Period:
Date:
13b. Could this trait
possibly be advantageous
in a specific environment?
Explain your answer.
584
Analysis of Human Traits, HASPI Medical Biology Lab 22b