BIOLOGY THIRTY
UNIT C2 – GENETICS
CHECK YOUR UNDERSTANDING
NAME:
UNIT C2 – GENETICS – CHECK YOUR UNDERSTANDING
Task
1
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10
Assessment Level (1-5 or CPT/NYC)
ALL ASSESSMENTS COMPLETED ON TIME?
____ YES
____ NO
DATE: ______________
**ASSESSEMENT LEVEL BREAKDOWN**
LEVEL 5
Addresses all complex curriculum outcomes correctly.
LEVEL 4
Addresses all complex curricular outcomes. One major component is incorrect or lacking
evidence of understanding.
LEVEL 3
Sufficient understanding of basic curricular outcomes. Understanding of complex outcomes is
weak.
LEVEL 2
With help demonstrates a partial understanding of some simpler outcomes. Limited
understanding of complex outcomes.
LEVEL 1
Little evidence of understanding of curricular outcomes despite help being offered.
**Some homework log assignments will be assigned a CPT or a NYC depending on the nature of the task.
BIO30: Unit C2  C.Y.U. Booklet
TASK #1 – Genetics Assignment – Monohybrid Crosses
Part 1: MONOHYBRID CROSSES
Alternate Gene Form
Parents
ORGANISM
CHARACTERISTIC
DOMINANT
RECESSIVE
MALE
FEMALE
tulip
Flower colour
Body hair
Mane length
Flower location
Plant height
Wing span
Eye colour
Tail shape
Branch pattern
Colour
yellow
bald
short
terminal
tall
normal
brown
curved
thick
green
blue
hairy
long
lateral
short
abdominal
blue
straight
sparse
yellow
Homo/rec
Hetero
Homo/dom
Hetero
Short
Hetero
Blue
Homo/rec
Homo/dom
Hetero
Hetero
Homo/dom
Long mane
Homo/rec
Short
Hetero
Hetero
Homo/dom
Hetero
Hetero
fruit flies
horses
corn
peas
beetles
human
monkeys
pine tree
algae
Using punnet squares, answer the following questions, you may need to refer to the
chart on the previous page to determine the dominant or recessive alleles.
1. A botanist crossed two “terminal flowered” corn plants. Could you accurately
predict the outcome of such a cross? Explain the reasons for your answer.
2. Two beetles (a normal winged male and a abnormal winged female) mate and
produce 70 F1 offspring. Thirty-three F1’s were normal winged and thirty-seven
were abnormal winged. What were the genotypes of the two parents involved?
Show the punnet square representing this cross.
3. A pet shop owner has a male monkey with a curved tail. He wishes to find out
whether the monkey is homozygous or heterozygous. Biologist “A” tells him to
mate the male to several straight tailed females. He believed that the results of
these crosses would answer the question. Biologist “B” says that the owner
must mate the male with a number of curve-tailed females to answer the
question. Which Biologist is correct? Explain your answer.
4. Two heterozygous yellow tulips are crossed. 84 seeds were produced by the
female plant. If all eighty-four germinated and grew, how many would be
expected to be yellow and how many would be expected to be blue?
5. A “bald” male fruit fly crossed with a “hairy” female fruit fly and all of the
offspring were bald.
a. What were the genotypes of the parents?
b. What results are expected if two F1 individuals were crossed to produce
and F2 generation?
Use the following information to answer the next three questions.
A study published in the journal Pediatrics indicates that breast-fed infants have
a substantially decreased risk of developing diarrhea compared with infants fed
formula. Another study reported that although a majority of infants harbor
populations of bacteria that would cause diarrhea in adults, breast-fed infants do
not get sick. The bacterium Clostridium difficile produces a toxin that irritates the
lining of the colon causing diarrhea. Breast milk contains a protein called
secretory component that binds to the toxin, thus causing the toxin to be
ineffective.
The human milk protein, secretory component, can be manufactured by
transgenic sheep. The following steps are necessary for producing transgenic
sheep.
1. The gene for secretory component is isolated and cloned into a vector.
2. Vectors carrying the gene are microinjected into fertilized sheep eggs,
which are then implanted into female sheep.
3. Heterozygous transgenic offspring are identified.
6. The heterozygous offspring described above are next mated to non-transgenic
sheep. If the allele for human secretory protein is s and the absence of the
human gene is S, the symbolic representation of the cross
a.
b.
c.
d.
ss
Ss
Ss
SS
X
X
X
X
ss
ss
SS
ss
7. The portion of offspring from the mating of a heterozygous transgenic sheep and
a non-transgenic sheep that are predicted to be heterozygotes is
_______________.
8. The heterozygous offspring are then mated and their homozygous transgenic
offspring are used for producing the milk product. Out of 220 offspring
produced from these crosses, how many offspring are predicted to be transgenic
homozygotes? _____________________________
9. A gardener has a pine tree with a “thick” branching pattern.
a. Would he know the tree’s phenotype? Explain
b. Would he know the tree’s genotype? Explain
10. John is brown eyed but he has a blue-eyed father. Both of Mary’s parents were
blue eyed. Using a punnet square answer the following:
a. What are the chances of John and Mary producing a blue-eyed child?
b. What are the chances of John and Mary producing a brown-eyed child?
11. A farmer had a stallion with a short mane. He bred the stallion to three different
mares (A,B,C). The results of the crosses are shown below:
Stallion
Stallion
Stallion
X Mare A (long mane) >>>>>> Colt #1 (short mane)
X Mare B (long mane) >>>>>> Colt #2 (long mane)
X Mare C (short mane) >>>>>> Colt # 3 (long mane)
Determine the genotypes of each organism.
12. Two tall pea plants are crossed and 282 seeds are produced. After planting, only
235 reach maturity with 175 of those being tall and 60 being short.
a. Do the results of such a cross meet expectations of a typical monohybrid
cross? Explain.
b. What are the genotypes of the two original parents?
Part 2: DOMINANT AND RECESSIVE ALLELES
1. Several black guinea pigs of the same genotype were mated and produced 29
black and 9 white offspring. What would you predict the genotypes of the
parents to be?
2. If a black female guinea pig is test crossed and produces at least one white
offspring, determine
a. the genotype and phenotype of the sire (male parent) that produced
the white offspring,
b. the genotype of this female
3. Heterozygous black (Bb) guinea pigs are mated to homozygous recessive (bb)
white. Predict the genotypic and phenotypic ratios expected from backcrossing
the black F1 progeny to
a. The black parent
b. The white parent
4. The lack of pigmentation, called albinism, in humans is the result of a recessive
allele (a) and normal pigmentation is the result of its dominant allele (A). Two
normal parents have an albino child. Determine the probability that
a. The next child is albino
b. The next 2 children are albino
c. What is the chance of these parents producing 2 children, 1 albino and
the other normal?
5. Short hair is due to a dominant gene L in rabbits, and long hair to its recessive
allele l. A cross between a short-haired female and a long-haired male produced
a litter of 1 long-haired and 7 short-haired bunnies.
a. What are the genotypes of the parents?
b. What phenotypic ratio was expected in the offspring generation?
c. How many of the 8 bunnies were expected to be long-haired?
Part 3: CODOMINANT ALLELES
1. When chickens with splashed which feathers are crossed with black-feathered
birds, their offspring are all slate blue (Blue Andalusian). When Blue Andalusians
are crossed among themselves, they produce splashed white, blue, and black
offspring in the ratio of 1 : 2 : 1, respectively.
a. How are these feather traits inherited?
b. Using any appropriate symbols, indicate the genotypes for each
phenotype
2. Yellow coat colour in guinea pigs is produced by the homozygous genotype CYCY ,
cream colour by the heterozygous genotype CYCW , and white by the homozygous
genotype CWCW . What genotypic and phenotypic ratios are matings between
cream-coloured individuals likely to produce?
3. The Palomino horse is a hybrid exhibiting a golden colour with lighter mane and
tail. A pair of codominant alleles (D1 and D2) is known to be involved in the
inheritance of these coat colours. Genotypes homozygous for the D 1 allele are
chest-nut coloured (reddish), heterozygous genotypes are Palomino-coloured,
and genotypes homozygous for the D2 allele are almost white and called
cremello.
a. From matings between Palominos, determine the expected Palomino :
non-Palomino ratio among the offspring
b. What percentage of the non-Palomino offspring in the part (a) will breed
true?
c. What kind of mating will produce only Palominos?
Part 4: LETHAL ALLELES
1. In the Mexican Hairless breed of dogs, the hairless condition is produced by the
heterozygous genotype (Hh). Normal dogs are homozygous recessive (hh).
Puppies homozygous for the H allele are usually born dead with abnormalities of
the mouth and absence of external ears. If the average litter size at weaning is 6
in matings between hairless dogs, what would be the average expected number
of hairless and normal offspring at weaning from mating between hairless and
normal dogs?
2. Thalassemia is a hereditary disease of the blood of humans resulting in anemia.
Severe anemia (thalassemia major) is found in homozygotes (TMTM) and a milder
form of anemia (thalassemia minor) is found in heterozygotes (TMTN). Normal
individuals are homozygous TNTN. If all individuals with thalassemia major die
before sexual maturity,
a. What proportion of the adult F1 from marriages of thassemia minors by
normals would be expected to be normal
b. What fraction of adult F1 from marriages of minors by minors would
expect to be anemic?
Part 4: MULTIPLE ALLELES
1. A multiple allelic series is known in the Chinese primrose where A (Alexandria
type = white eye) > an (normal type = yellow eye) > a (Primrose Queen type =
large yellow eye). List all of the genotypes possible for each of the phenotypes in
this series.
2. The coat colours of many animals exhibit the “agouti” pattern, which is
characterized by a yellow band of pigment near the tip of the hair. In rabbits, a
multiple allelic series is known where the genotypes EDED and EDe produce only
black (nonagouti), but the heterozygous EDE produces black with a trace of
agouti. The genotypes EE or Ee produce full colour, and the recessive genotype
ee produces reddish-yellow. What phenotypic and genotypic rations would be
expected in the F1 from the cross
a. EDED X Ee
b. EDe X ee
Part 5: MIXED PRACTICE
1. The genotypes of the parents to whom this Punnett square applies are
a. Heterozygous B and homozygous A
b. Heterozygous O and homozygous A
c. Homozygous B and heterozygous A
d. Heterozygous B and heterozygous A
2. The genotype of the parent plant with long V-shaped lines and of the parent
plant with short V-shaped lines are, respectively,
a. VhVl and vv
b. Vhv and Vlv
c. VhVl and Vlv
d. Vhv and VlVl
3. Using the numbers above, match these descriptions and symbols with the term
below which they apply
_____________ _________________ _________________ _____________
gene
allele
phenotype
genotype
4. What are the genotypes for coat colour of two horses that are predicted to
produce offspring in a 1 : 1 genotypic ratio?
a. Tt and tt
b. Tt and Tt
c. Tobiano and tobiano
d. Tobiano and not tobiano
5. The likely genotypes of the P1 plants for these two crosses is represented in row
ROW
A
B
C
D
P1 genotypes
Scarlet-red-flowered
offspring
P
R R x rr
RPRP x rr
RPr x RPr
RPRP x RR
P1 genotypes
Orange-flowered
offspring
RR x rr
RR x rr
Rr x Rr
RPR x Rr
6. Which of the following phenotypes is the predicted flower colour of M. jalapa
with the genotype RPR?
a. Yellow
b. Orange
c. Crimson
d. Scarlet-red
7. Which of the following rows indicates the relationship between IA and IB alleles
and the relationship between IA and I alleles for the blood type gene?
ROW
A
B
C
D
Relationship between
IA and IB
Codominant
Codominant
Dominant-Recessive
Dominant-Recessive
Relationship between
IA and i
Codominant
Dominant-Recessive
Codominant
Dominant-Recessive
TASK #2 – Dihybrid Crosses
BIO30: Unit C2  C.Y.U. Booklet
1.
In guinea pigs, black coat color is dominant to white and short hair is dominant to long.
a. What is the genotype of a heterozygous black short guinea pig?
b. What gametes can this guinea pig produce?
c. Outline a cross and state phenotypic ratios between a heterozygous black short guinea pig and a white long guinea pig.
2.
When mountain sheep of poor eyesight are crossed with pure sharp-sighted sheep all of the offspring are sharp sighted. When
two hybrid white coated sheep are crossed, 3 offspring are white and 1 is grey. Outline a cross between a heterozygous sharp
sighted, white ram and a poor-sighted grey ewe.
3.
In shorthorn cattle, normal hoof is dominant to abnormal hoof. Red and white coat color genes show co-dominance so that
heterozygous individuals have a roan coat color. Indicate the phenotypic ratio from crossing two heterozygous normal hoofed,
roan shorthorn cattle.
4.
There are two methods used for blood typing. The ABO blood group and the Rh factor where Rh+ is dominant over Rh-. A
woman who is A+ and a man who is O+ mate and have an O- child.
a. Indicate the genotypes of the two parents.
b. What is the probability there next child will be a girl who is O+?
c. What is the probability there next two children will both be girls, one being A- and the other being O+?
5.
In squash plants white and yellow are fruit colors, and disk and sphere are fruit shapes. The cross of a white disk and a yellow
sphere produces an F1 of all white disks. The F2 generations consists of 9 White disks, 3 white spheres, 3 yellow disks and 1
yellow sphere. Use this information to establish the inheritance pattern of both genes, show all work.
6.
The gene R produces a rose comb, in chickens. Another gene, P, which is located on a different chromosome, produces a pea
comb. The absence of the dominant rose comb gene and pea comb gene (rrpp) produces birds with single combs. However,
when the rose and pea comb genes are present together they interact to produce a walnut comb (R_P_). Indicate the
phenotypes of the parents and give the phenotypic ratios of the F1 generation from the following crosses:
a. rrPP x RRpp
b. RrPp x RRPP
c. RrPP x rrPP
d. RrPp x RrPp
7.
In four o’clock plants, when pure red flowered, tall stemmed plants are crossed with pure white-flowered, short stemmed plants,
all of the F1’s produced are pink flowered and medium stemmed. What are the phenotypic results when a pink tall flower is
crossed with a white medium flower?
8.
In humans deafness can be a result of two different recessive genes. Middle ear deafness results when the ossicles do not
develop properly thus inhibiting conduction of sound. In the other type of deafness, Auditory nerve deafness, the recessive
genotype causes underdevelopment of the auditory nerve allowing no neural messages to the brain. Both types of deafness are
only expressed in homozygous form. Each gene has a dominant allele that expresses “normal hearing”. When two normal (not
deaf) parents who are both “carriers” of both recessive genes mate, what is the probability that their offspring will be normal?
Deaf?
9.
In pea plants, round seeds are dominant over wrinkled and purple flowers are dominant over white. What are the expected
phenotypic ratios from a cross between
a. A heterozygous round purple and a pure breeding plant of the same phenotype
b. A heterozygous round purple and a pure breeding plant of different genotype
10. For mice, the allele C produces color. The genotype cc causes albinism and is epistatic to the gene B=black and b=brown
a. What are the possible genotypes of an albino mouse?
b. Two black mice (with the same genotype) are crossed and produce two offspring, one is brown and the other is albino.
What are the genotypes of the two black mice?
c. Outline a possible cross between a heterozygous black mouse and pure-breeding albino mouse.
TASK #3 – Genetic Probability
BIO30: Unit C2  C.Y.U. Booklet
1. A mother with brown eyes and a father with blue eyes have a child with blue eyes.
a. Indicate the genotypes of each parent.
b. What is the probability that their next child will be a girl with brown eyes?
c. What is the probability that their next two children will both be boys, one with brown
eyes and the other with blue eyes?
2. In horses, a long coat is dominant over a short coat. Black coloring is dominant over brown
coloring. A male, heterozygous for both traits, crosses with a short coated, hybrid black female.
a. Outline the results of this cross using a Punnett Square.
b. What is the probability that their next child will be a girl with a black long coat?
c. What is the probability that their next 2 children will both be boys, one with a short
brown coat, the other with a long black coat.
3. There are two methods used for blood typing. The ABO blood group and the Rh factor where
Rh+ is dominant over Rh-. A women who is B- and a man who is O+ have a child who is O-.
a. Identify the genotypes of each parent.
b. What is the probability that their next child will be a boy who is O+?
c. What is the probability that their next two children will both be girls, one being B+
and the other being O+?
TASK #4 – Pedigree and Probability Review
BIO30: Unit C2  C.Y.U. Booklet
PROBABILITY QUESTION #1
X-linked dominant hypophosphatemic rickets, also known as vitamin d-resistant rickets
can cause bone deformity and result in a short stature or bow leggedness. A mother
suffering from this disorder has a father who was normal and mother who was bowlegged. She has a child with a man who is normal.
a. State the potential genotypes of the child using a Punnett square.
b. What is the probability that the mother and father will have a female child who
suffers from rickets?
c. If one of their children is an affected male, what is the probability that this individual
will have two children, one normal male and one female with rickets, when he mates
with a female who is heterozygous for the disorder?
PROBABILITY QUESTION #2
In humans, the trait for freckles dominates over regular skin tone. Thomas’ paternal
grandfather had freckles due to a heterozygous genotype, whereas his paternal
grandmother had a regular skin tone. Thomas’ father also had freckles, which his
mother who has perfectly normal skin found quite attractive. Thomas has freckles and
his wife, Jane, does not.
A. Identify the genotypes of Thomas and Jane.
B. What is the probability that Thomas and Jane will have a daughter with freckles?
C. What is the probability that Thomas and Jane will have three children, 1 boy with
freckles, and 2 girls with regular skin tone?
PEDIGREE #1 – Cystic Fibrosis
A. Identify the inheritance pattern of Cystic Fibrosis.
B. Calculate the probability that individuals II.6 and II.7 will have a daughter with
Cystic Fibrosis.
PEDIGREE #2 – Huntington’s Disease
A. Identify the inheritance pattern of Huntington’s disease.
B. Calculate the probability of individuals I.1 and I.2 having a son with Huntington’s
disease.
PEDIGREE #3 – Vitamin D Resistant Rickets
A. Identify the inheritance pattern of vitamin D resistant rickets.
B. Calculate the probability that individuals II.3 and II.4 will have a daughter with
vitamin D resistant rickets.
PEDIGREE #4 – Hemophilia
A. Identify the inheritance pattern of hemophilia.
B. Calculate the probability of individuals I.1 and I.2 having a daughter with
hemophilia.
BIO30: Unit C2  C.Y.U. Booklet
TASK #5 – Gene Mapping
1. The crossover frequency between genes E and F is 6%, between E and G is 10% and
between F and G is 4%. Construct a map of these genes and indicate the order of
the genes.
2. In 1911, Thomas Hunt Morgan collected the following crossover gene frequencies
while studying Drosophila. Bar-shaped eyes are indicated by the B allele, and
carnation eyes are indicated by the allele C. Fused veins on wings (FV) and scalloped
wings (S) are located on the same chromosomes. Construct a gene map and indicate
the order of these genes.
Gene Combinations
FV/B
FV/C
B/C
B/S
FV/S
C/S
Recombination Frequency
2.5%
3.0%
5.5%
5.5%
8.0%
11.0%
3. The following chart shows the crossover frequencies for some genes on an
autosome of organism Z. Construct a chromosome map, indicate the order of the
genes, and determine the map units from R & S.
Genes
P&Q
P&R
P&S
Q&R
Q&S
Crossover Frequency
5%
8%
12%
13%
17%
4. For a series of experiments, a linkage group composed of genes W, X, Y and Z was
found to show the following gene combinations. (All recombinations are expressed
per 100 fertilized eggs). Construct a gene map and indicate the order of the genes.
Genes
W
X
Y
Z
W
–
5
7
8
X
5
–
2
3
Y
7
2
–
1
Z
8
3
1
–
5. In Drosophila, the gene for eye color (pr), wing shape (vg), and body color (eb) are all
found on the same chromosome. The following crossover frequencies for these
genes were determined by experiments.
Genes
pr & vg
pr & eb
vg & eb
Crossover Frequency
12.5%
6.0%
18.5%
What is the correct sequence of the genes pr, vg and eb on the chromosome?
6. Construct a gene map given the following information:
1. Determine the number of map units for each gene
2. Construct a gene map and determine the order of the genes.
Genes
A&B
A&C
C&D
A&F
F&B
D&F
Crossover Frequency
24.0%
8.0%
2.0%
16.0%
8.0%
6.0%
Map Units
TASK #6 – Genetics Long Answer Practice
BIO30: Unit C2  C.Y.U. Booklet
11Adrenoleukodystrophy (ALD) is a rare disease of the central nervous system. ALD is
characterized by the accumulation of very-long-chain fatty acids in the white matter of the
brain and in the adrenal glands. These fatty acids cause the myelin sheath on nerve fibres
within white matter of the central nervous system to degenerate. Symptoms of this
degeneration become more severe as more and more fatty acids accumulate. Symptoms
start with tantrums and other behavioural problems; then motor function, speech, and
hearing are impaired; and finally blindness, mental deterioration, and death occur. ALD
can be partially diagnosed by abnormally high ACTH levels in the blood.
Hereditary diseases have diverse causes. For example, the disease mutation may be
dominant or recessive, or the mutated gene may be present on the X chromosome or on
an autosome. In some cases, similar diseases are caused by mutation in two different
genes. One such case is ALD, where one gene is autosomal and the other is X-linked. In
both forms of inheritance, the disease mutation is recessive. Scientists continue to
research the causes of ALD. The X-linked recessive form of ALD can be diagnosed
prenatally.
ALD has not been treated successfully; however, bone marrow transplants and a diet
restricted in a very-long-chain fatty acids have shown promise. One dietary substance,
oleic acid (Lorenzo’s oil), has been successful in normalizing levels of very-long-chain fatty
acids in improvement in symptoms, particularly in brain degeneration. Lorenzo’s oil did
produce positive clinical results when the treatment of patients began before neurologic
symptoms were present.
1. Describe one piece of evidence obtained from the analysis of a pedigree chart
that could be used to determine whether the mode of inheritance of a human
genetic disorder is X-linked or autosomal and one piece of evidence that could
be used to determine whether it is recessive or dominant. Construct a pedigree
of four generations that clearly illustrates one of the two types of inheritance of
ALD. Clearly label where your pedigree shows evidence of X-linked recessive or
autosomal recessive inheritance.
TASK #7 – Genetics Long Answer Practice 2
BIO30: Unit C2  C.Y.U. Booklet
Larry and Danny Gomez, two boys known as “Wolf Boys,” have made the circus their adoptive family.
Both boys perform as trampoline acrobats, and Danny also does motorcycle stunts. The boys have a
condition called congenital hypertrichosis (CH), which is a very rare X-linked dominant inherited
condition. CH is characterized by the growth of dark hair over the body, particularly on the face and
upper torso in males. The palms of the hands, soles of the feet, and mucus membranes are not
affected by this condition. A press release about the circus stated that Larry and Danny have travelled
to many countries in search of a cure. When asked about the search for a cure in an interview by David
Staples of The Edmonton Journal (May 14, 1997), Larry said, “I’d never take it off. I’m very proud to be
who I am.” Outside the circus, the boys enjoy activities typical of most boys their age. Danny likes to
play video and board games, and Larry is interested in science and is taking astronomy by
correspondence.
Researchers continue to investigate the process of hair growth and the cuases of hair distribution at the
molecular level. The relevant molecules are expected to act on hair follicels. Hair follicle distribution in
humans is primarily a hormone-dependent secondary sex characteristic. In addition to searching for a
cure for CH, research in this area may also have significant applications in the treatment of acquired or
inherited baldness.
The incidence of CH is very rare: only about 50 affected individuals have been reported since the
Middle Ages. The incidence of this condition is considerable higher in a small Mexican village that it is
in the rest of the human population. In 1984, researcher Macias-Flores studied CH in a large, fivegeneration Mexian family and found 19 individuals with CH. A partial pedigree showing the sampled
individuals from the Macias-Flores study is shown below.
1. Identify the genotypes for individuals II-4, II-5, III-11, III-12, IV-6, IV-7, IV-8, and
IV-9 in one of the lines of inheritance on the pedigree. (Provide a key for the
allele symbols you use.) Construct a Punnett square to predict the probability of
individuals III-11 and III-12’s next child being a male with CH. Explain why more
females than males inherit CH in generation III.
BIO30: Unit C2  C.Y.U. Booklet
TASK #8 – Genetic Counseling
It Says, I Say, and So…
It says  What does the article say about genetic counseling?
I say  How do you feel about genetic counseling?
and So  What conclusions can you draw on genetic counseling based on your opinions and what you have read in the
article?
It Says…
I Say…
and So…
BIO30: Unit C2  C.Y.U. Booklet
TASK #9– Genetic Therapy
Read the article shared with you on Google Docs and complete the assignment below.
PROS, CONS, & QUESTIONS
ISSUE:
PROS
QUESTIONS YOU ARE LEFT WITH:
CONS
BIO30: Unit C2  C.Y.U. Booklet
TASK #10 – PKU: A Genetic Disorder
GETTING THE GIST
Article Title:
WHO?
WHAT?
WHERE?
WHEN?
WHY?
HOW?
GIST (article summarized in 20 words or less)