Download Biology Homework: Genetics

Biology Genetics Unit HW Packet #1
Name: _________________________
Hour: _____
Due: ___________
HW #1
Coin Toss
Lab
HW #2
Nature vs.
Nurture
Table
Analysis
______
______
Data Table
Hypothesis
Analysis
____
____
____
______
Monohybrid
WS #1
Completion
Monohybrid
WS #2
Completion
______
______
Genetic
Disorder
Completion
______
Practice
Extra practice
______
_____
5 points possible
1
SS
Ss
Ss
ss
1
25
2
1
50
25
3
1
75
25
What gene combinations are expected from a cross when
you use a Punnett square.
Mathematically
What gene combinations seen from actual crosses.
Chance
Heads = dominant normal skin color
Tails = recessive albino skin color
There is a 50% chance for either heads or tails. This is the same for either
dominant or recessive gene
Chromosomes (and genes) come in pairs. You need to use 2 coins to represent this situation.
Sperm fertilizing egg cell creating a diploid zygote; dad’s chromosomes with mom’s chromosomes.
No
In any cross, expected and observed.
data may not exactly agree. Ex pected based on Punnett square; observed based on actual crosses.
More results should come closer to what is expected based on Punnett square.
No
In any cross, expected and observed.
data may not exactly agree. Ex pected based on Punnett square; observed based on actual crosses.
Yes
Yes
No
This sample size is too small to give accurate data.
By using a larger sample size, the observed results should come closer to the expected results.
48
37
36
(48 x .75)
Ye s
11
12 (48 x .25)
Ye s
No. 15 people should have 3-4 albino (11 should be normal).
Observed resul ts show 2 albinos (13%) and 13 normal (87%).
Ye s. When looking at all nine families, we expect ¾ (75%) to have
normal skin; observed 77%. Expect to see ¼ (25%) to have albinism;
observed 23%. A larger sample size should bring observed results
closer to the expected results.
Coin Toss Lab Questions
No
1.
Can you control the outcome of a single coin toss? (yes or no)
2.
A heterozygous parent (Ss) mates to produce a single offspring. What is the chance of
the offspring inheriting this parent’s dominant gene?
?
?
S
s
S?
?s
S?
?s
Chance of inheriting
dominant gene = 2/4
In Table 18-2, are the numbers in columns B and C exactly the same? (yes or no) No
3.
Should they be? (yes or no) No (but they should be close)
Explain (be brief). Expected results are calculated mathematically.
Observed results are those that occur in actual crossings. These
should be close; more events should yield closer results.
4.
Matching: Match the definition on the left with the letter of the corresponding item on
the right. (some items will not be used)
G
What a single coin represented.
A.
3 Normal: 1 Albino
____E___
Genotype for albino skin
B.
Sample size
____A___
Phenotypic ratio expected
from the cross: Ss X Ss
C.
a diploid body cell
____D___
Genotypic ratio expected
from the cross Ss X Ss
D.
1SS : 2Ss : 1ss
____B___
Will determine how close observed
and expected values are to one
another
E.
ss
F.
G.
4 normal
a haploid cell
H.
2SS : 2Ss
HW # 2 – Plant Phenotypes “Nature vs. Nurture”
Background:
Is heredity or environment more important in determining the kinds of traits that appear in
offspring? For years scientists and psychologists have argued the relative importance of genes
and how you are raised. Many studies of twins raised in different households have yielded
surprising data. We will study the effect of genetics (the alleles you possess) and environment
(the growing conditions) on a batch of corn seeds. The corn we will grow has two alleles, Green
is dominant G = green and albino is recessive g = albino. The seeds you will use from parents
plants were both heterozygous (Gg).
Hypothesis: By using a Punnet Square determine the possible genotypes of seeds produced
by plants heterozygrous for green (Gg).
G
GG
g
Gg
3 out of 4
will be green
G
g
Gg
gg
1 out of 4
will be albino
Procedure A:
1. The teacher will prepare the sample of corn and albino plants into a classroom
batch. .
2. One dish will be marked “dark” and placed in the dark to grow. The other will be
marked “light” and placed directly in the light.
3. After 6 days, observe the plants that have grown from seeds in the batches.
Note how many in each dish are green or albino (white or yellow). Combine your
results totals with those of your classmates to complete Table 1.
4. After your observations, place both batches in a light environment for several
days. Moisten again if necessary.
Data Table 1:
CLASS RESULTS OF FIRST OBSERVATION
1.
Plants in Dark
# of green # of albino % of albino
Plants in Light
# of green # of
% of albino
albino
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Procedure B:
1. After 2 days, observe the plants. Note how many in each dish are green or albino
(white or yellow).
2. Combine your results (totals) with those of your classmates to complete Data
Table 2.
Data Table 2:
CLASS RESULTS OF SECOND OBSERVATION
2.
Plants in Dark
# of green # of albino % of albino
Plants in Light
# of green # of
% of albino
albino
BACKGROUND
Corn plants require two factors in order to produce chlorophyll (green pigment in plants that
attracts sunlight). They must have the proper gene combination (at least one dominant “G”
gene) and also be exposed to light.
When doing any experiment, the more data or results you can gather, the more reliable your
conclusions should be.
Analysis:
1. Explain why the percentage of albino plants in the dark in the first observation did not
agree with expected results obtained from the Punnett Square.
Not only do corn plants require the proper gene combination but they also require
light to produce chlorophyll. The corn plants in the first observation did not have
light so were unable to produce chlorophyll therefore our expected results from the
Punnett Square did not agree with what was observed.
2. What happened to the percentage of white plants when those in the dark were placed in
the light for several days?
The percentage of white plants went down. More plants turned green.
3. Does the percentage of albinos at this point agree or come close to the expected
percentage of albino’s in your Punnett Square? Explain.
The observed percentage of albinos should come closer to the expected number
because the plants were exposed to light allowing the chlorophyll to be produced.
4. What happened to the percentage of albino plants when those in the light remained in
the light for several more days? Explain.
The percent albino should not change. If a plant is albino it contains 2 recessive
genes. They will always show that phenotype.
5. Explain how it is possible for environment to influence or temporarily change the
expression of a gene. Plants grown in the dark have the ability to be green, but
the environment didn’t allow them to use the genes for chlorophyll. There is
no need for chlorophyll in the dark (no photosynthesis can take place).
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Monohybrid Worksheet #1
A.
In a certain plant, tall stems are dominant to short stems. A farmer crosses a
short-stemmed flower with a homozygous long-stemmed flower. Complete each of the
seven steps.
1.
Dominant = long stem
Recessive = short stem
2.
Long = L
Short = l
3.
Mom = ll
4.
l
L
l
l
5.&6.
L
L
7.
Dad = LL
L
l
Ll
Ll
Ll
Ll
What is the chance of an offspring being tall ?___4/4____________
What is the chance of an offspring being short?__0/4____________
How many genotypes exist?________one___(Ll)____________
How many phenotypes exist?________one - long stemmed________
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B.
In a certain plant, yellow fruit is dominant to white fruit. A hybrid plant with
yellow fruit is crossed with a plant that has white fruit.
1.
Dominant = yellow fruit
Recessive = white fruit
2.
Y = yellow
y = white
3.
Mom = Yy
Dad =
4.
Y
5.&6.
yy
y
y
y
y
Y
y
Y
y
y y
Y
y
y y
y
7.
What is the chance of an offspring having yellow fruit?____2/4______
What is the chance of an offspring having white fruit?____2/4_______
How many genotypes exist?___2_______
What are they?_____
Y
y
___
y y
How many phenotypes exist?___two
What are they?___Yellow and white
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C. Blue larkspur flowers are dominant over white ones. The male flower contains
heterozygous genes. The female flower contains only blue genes. Perform the seven
steps.
1.
Dominant = yellow fruit
Recessive = white fruit
2.
B = blue
b = white
3.
Dad = Bb
4.
Mom = BB
b
B
5.&6.
B
B
B
b
B
BB
Bb
B
BB
Bb
y
7.
What fraction of their kids should have white flowers?____0/4______
List the different genotypes that their kids can have:_____BB,
List the phenotypes their kids can have:_____only
Bb___________
blue flowers______________
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Monohybrid Worksheet #2
A. Tallness is dominant over dwarfism in pea plants. If one plant is homozygous tall
and the other plant is heterozygous tall, predict the possible genotypes and
phenotypes for this cross.
1.
Dominant = tall
Recessive = dwarfism
2.
T = Tall
t = dwarfism
3.
Mom = TT
4.
Dad = Tt
T
T
T
t
5 & 6.
T
T
t
7.
T
TT
TT
Tt
Tt
What is the chance of an offspring being tall?_____4/4__________
What is the chance of an offspring being short?______0/4________
How many genotypes exist?______two_______________________
What are they?________TT and Tt_________________
How many phenotypes exist?______one_______________________
What are they?__________Tall___________________
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B. In seals, the gene for the length of the whiskers has two alleles. The dominant
allele (W) codes long whiskers and the recessive allele (w) codes for short
whiskers. What percentage of offspring would be expected to have short
whiskers from the cross of two long-whiskered seal s, one that is homozygous
dominant and one that is heterozygous.
1.
Dominant = long whiskers
Recessive = short whiskers
2.
W = long
w = short
3.
Mom = WW
Dad = Ww
4.
W
W
w
W
5 & 6.
W
7.
W
W
WW
WW
w
Ww
Ww
What is the chance of an offspring having short whiskers?_____0/4_______
What is the chance of an offspring having long whiskers?______4/4_____
How many genotypes exist?__________two________________
What are they?_______WW and ___
Ww
How many phenotypes exist?_________One__________________
What are they?__________long whiskered___________________
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C. Brown eyes are dominant over blue eyes. Explain three different genotype combinations
that the parents could have to get blue eyed offspring.
bb x bb
b
b
b
Bb x bb
b
B
b
bb
bb
b
Bb
bb
bb
bb
b
Bb
bb
Bb x Bb
B
b
B
b
Bb
Bb
Bb
bb
D. In mice, black fur is dominant to white fur. You find a black mouse. How could you
determine whether this mouse is homozygous dominant or heterozygous?
Cross a black mouse with a white mouse that you know is purebred
recessive. If any offspring end up white, you will know that the black
mouse is hybrid. You may need to breed several litters of mice from
the cross. Observed results come closer to expected results the more
offspring that are born.
B
B
B
b
b
Bb
Bb
b
Bb
bb
b
Bb
Bb
b
Bb
bb
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My assigned genetic disorder is ____________________
Genetic disorders are caused by changes to DNA. These changes are either really small and
involve only one tiny piece of DNA or really large and result in an entire missing chromosome.
Either way, the affect on how someone lives their life can be dramatic. Additionally, some
genetic disorders mean that a person may require someone to care for them their entire life.
Remember, your audience for this brochure is either someone who has just been diagnosed as
having this genetic disorder or someone who is the parent or child of someone who has been
diagnosed. Your teacher will show you examples of good brochures so you can be familiar with
the qualities you should include in your brochure.
GENETIC DISORDER PROJECT ASSIGNMENT #1 –
1. What are some of the symptoms of your assigned genetic disorder? Answer in complete
sentences. Be sure to attach a copy of your references to the homework packet.
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Practice Problems for Monohybrid Quiz
I. Cross a mother with purebred blonde hair and a father with hybrid brown hair.
1. Dominant = Brown
Recessive = Blonde
2. B = Brown
b = blonde
3. Mom = bb
Dad = Bb
4.
b
B
b
5 & 6.
b
B
b
b
Bb
bb
b
Bb
bb
7. Tell me the genotypes present, probability of each, and the percent of offspring with each
phenotype. Bb = 2/4 (50% - brown); bb = 2/4 (50%-blonde)
II. Diabetes insipidus, a failure to adequately concentrate urine leading to gallons of urine
production a day is caused by mutation in the aquaporin-2 gene that causes there to be no
response to Anti-diuretic hormone (ADH) in the kidneys.(http://www3.ncbi.nlm.nih.gov/htbinpost/Omim/dispmim?125800) It is passed down in an autosomal dominant manner. Cross a
homozygous dominant male with a heterozygous female.
1. Dominant = Diabetes insipidus
Recessive = healthy
2. D = diabetes
d = healthy
3. Dad = DD
Mom = Dd
4.
D
D
5 & 6.
D
D
D
DD
DD
D
d
d
Dd
Dd
7. Tell me the genotypes present, probability of each, and the percent of offspring with
each phenotype. DD = 2/4 ; Dd = 2/4 ; 100% offspring with diabetes insipidus
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