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Biology Genetics Unit HW Packet #1

Name: _________________________ Hour: _____

Due: ___________

HW #1

Coin Toss Table ______

Lab Analysis ______

HW #2

Nature vs. Data Table ____

Nurture Hypothesis ____

Analysis ____

Monohybrid Completion ______

WS #1 ______

Monohybrid Completion ______

WS #2 ______

Genetic

Disorder

Completion ______

Practice Extra practice _____

5 points possible

SS Ss

Ss ss

1

1

2

25

25

50

1

3 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

Chromosomes (and genes) come in pairs. You need to use 2 coins to represent this situation.

There is a 50% chance for either heads or tails. This is the same for either

dominant or recessive gene

Sperm fertilizing egg cell creating a diploid zygote; dad’s chromosomes with mom’s chromosomes.

No

No

No

data may not exactly agree. Expected based on Punnett square; observed based on actual crosses.

In any cross, expected and observed.

More results should come closer to what is expected based on Punnett square.

In any cross, expected and observed.

data may not exactly agree. Expected based on Punnett square; observed based on actual crosses.

Yes

Yes

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)

Yes

11

12 (48 x .25)

Yes

No. 15 people should have 3-4 albino (11 should be normal).

Observed results show 2 albinos (13%) and 13 normal (87%).

Yes. 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

1. Can you control the outcome of a single coin toss? (yes or no) 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?

3. In Table 18-2, are the numbers in columns B and C exactly the same? (yes or no)No

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 C. a diploid body cell

from the cross: Ss X Ss

____D___ Genotypic ratio expected D. 1SS : 2Ss : 1ss from the cross Ss X Ss

____B___ Will determine how close observed E. ss and expected values are to one another F. 4 normal

G. a haploid cell

H. 2SS : 2Ss

S s

?

?

S?

S?

?s

?s

Chance of inheriting

dominant gene = 2/4

2

g

g

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).

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 Plants in Light

# of green # of albino % of albino # of green # of

albino

% of albino

GG

Gg

Gg

gg

G

G

3 out of 4

will be green

1 out of 4

will be 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 Plants in Light

# of green # of albino % of albino # of green # of

albino

% of 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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l l

L L

l l

L

L

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 Dad = LL

4.

5.&6.

7. 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________

Ll

Ll

Ll

Ll

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Y y y y

Y y

y

y

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 = yy

4.

5.&6.

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

Y y

y y

Y y

y y

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B b

B B

b B

B

B

y

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 Mom = BB

4.

5.&6.

7. What fraction of their kids should have white flowers?____0/4______

List the different genotypes that their kids can have:_____BB, Bb___________

List the phenotypes their kids can have:_____only blue flowers______________

BB

Bb

BB

Bb

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T T T t

T

T

T

t

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 Dad = Tt

4.

5 & 6.

7. 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___________________

TT

TT

Tt

Tt

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W W W w

W W

W

w

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 seals, 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.

5 & 6.

7. 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 ___ W w

How many phenotypes exist?_________One__________________

What are they?__________long whiskered___________________

WW

WW

W w

W w

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b b

b

b

B b

b

b

B b

b

B

B B

b

b

b

b

B b

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 Bb x bb Bb x 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.

bb

bb

bb

bb

Bb

bb

Bb

bb

Bb

Bb

Bb

bb

Bb

Bb

Bb

Bb

Bb

bb

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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b b b B

b

b

B b

D D D d

D

D

D d

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.

5 & 6.

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/htbin-

post/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.

5 & 6.

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

Bb

bb

Bb

bb

DD Dd

DD Dd