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Lesson #1 Segregation
AGRO 315 Genetics
Questions due Sunday, August 28, 2011
1. Insert the number of the best fitting term in each of the blanks below.
1. allele
7. heterozygous
13. monohybrid
2. dominant
8. homozygous
14. recessive
3. egg
9. gene
15. reciprocal cross
4. F1
10. generation
16. segregation
5. F2
11.genotype
17. sperm
6. F3
12. phenotype
18. zygote
_____ the particular traits of an individual, could be appearance, taste, smell, sound or
any other trait that can be assessed that results from both the individuals genetic
makeup and their environment.
_____ offspring generated from crossing two individuals that vary by one trait.
_____ the genetic makeup of an individual with respect to the genes that control a
particular trait.
_____ an alternative version of a particular gene
_____ description of an individual that has two of the same versions of a particular gene
in each of their somatic cells.
_____ description of an individual that has one copy of one version of a gene and one
copy of an alternative version in each of their somatic cells.
_____ female gamete
_____ crossing two F1s produces this generation.
_____ term used for the hereditary units that were first described by Mendel as being
some type of particle that controls traits and maintains it’s identity as it is passed from
parent to offspring.
________ crossing a true breeding belted male pig with a true breeding solid female to
verify the results obtained from crossing a true breeding belted female and a true
breeding solid male pig.
Aa Aa Aa Aa
Aa Aa Aa
Aa Aa
Aa Aa
Aa
Aa
Aa
Aa
Aa
A
a
A
a
2. a. The cells depicted in cellular division above that have two copies of the A,a
gene are called:
b. The cells depicted above that have one copy of the A,a gene are called:
c. The overall process depicted above is called:
3. Refer to Mendel’s Monohybrid cross data at the end of this document
a. Draw a diagram to explain why Mendel got 6022 seeds with yellow
cotyledons and 2001 seeds with green cotyledons in the F2 of this
experiment.
b. What is the probability that a yellow cotyledon F2 seed will grow into a plant
that will self and produce both yellow and green progeny?
4. Use the image of these ears of corn to answer the following questions.
a. The seeds on these ears were a result of self pollinating a single corn plant.
Explain how one would perform this cross to insure that every seed produced
was a result of the self pollination. Use the terms “gametes” and “zygote” in
your explanation.
Below is a data set obtained from a genetic experiment in which the only trait
observed was corn kernel color (yellow vs. purple).
Parent 1
genotype
Parent 2
genotype Purple
Yellow
Offspring Offspring
yellow
X yellow
0
492
purple
X yellow
235
241
purple
X purple
452
139
b. Fill in the blanks above in the genotype columns with the appropriate
symbols. Use any letter you wish but use the upper case to represent the
dominant allele.
c. Which trait is the mutant trait in corn (purple or yellow seeds)? Explain why
you came up with this conclusion.
d. These results do not exactly fit the even ratios predicted. Why not?
e. The ear you have is a result of a self pollination. What was the genotype of
the parent? What information supports this conclusion?
5. Corn plants grown in Nebraska now have traits derived from genetic
engineering that were never found in corn before. These traits result from new
genes being added to a chromosome in the corn plant. These new genes are
called transgenes and the genetically engineered plants are transgenic. We will
study this genetic engineering process in detail later in the course.
The PAT gene gives the corn plant resistance to a weed killing herbicide called
Liberty by telling the plant how to make an enzyme that chops up the herbicide
molecule before the molecule damages the plant.
PAT
gene
Liberty degrading enzyme
Liberty herbicide
molecules
Plants that have one chromosome per cell with the PAT are called hemizygous.
f. List a person in your family that would be hemizygous for genes on the X
chromosome:
g. List a person in your family that could be heterozygous for genes on the X
chromosome.
h. Liberty herbicide is labeled for application at a rate that will allow corn plants
with the PAT gene to be resistant. If the PAT gene is dominant this would
mean…
_____ Plants homozygous and hemizygous for the PAT gene would have the
same level of resistance at the labeled rate of the Liberty herbicide.
____ Homozygous plants would be more resistant and hemizygous plants would
have less resistance at the labeled rate of the Liberty herbicide.
5. continued.
i.
The seeds pictured above are the F2 generation since hybrids farmers grow
are F1s. If you planted these seed and they developed into plants, what
generation will the plants be?
___ F1
___ F2 ___ F3
j. If the plants that grow from the seeds are sprayed with a little bit of Liberty
herbicide, and some were not sprayed, the sprayed plants might show a different
phenotype because…
___ environment is changed by Liberty application
___ genotype is changed by Liberty application
j. Explain how you could determine if the F1 parent of your seeds was
Homozygous for the PAT gene, hemizygous, or nontransgenic.
6. Blood tests reveal a person’s cholesterol levels.
a. Having high levels of “bad cholesterol” or LDL is a result of a person’s….
1. genotype
2. environment
3. genotype and environment
b. Tangier disease is very rare. One reason is because the defective version of
the ABC1 gene causing the disease is not very common. The other reason is
that people will only have the disease if they lack a normal version of ABC1 in
their cells. Therefore people with Tangier disease must be..
1. homozygous
2. heterozygous
A
3. either homozygous or heterozygous
?
B
?
C
Assume the pedigree shows the occurrence of Tangier disease (filled symbols)
c. What is the probability that ‘C’ will have Tangier disease?
d. What is the probability that ‘C’ will be a carrier of the disease allele but be
normal?
e. If ‘C’ has Tangier disease, what would be the chance of a sibling also having
the disease?
f. If the female parent of ‘B’ is homozygous dominant, what is the chance that ‘B’
will have the disease?
Mendel’s Result from Crossing Peas
Trait
Cross
F1
F2 number
F2 ratio
Seed form
Round X Wrinkled
All round
5474 Round
1850 Wrinkled
2.96 to 1
Cotyledon
Color
Yellow X Green
All yellow
6022 Yellow
2001 Green
3.01 to 1
Seed coat
Color*
Gray X White
All gray
705 Gray
224 White
3.15 to 1
Pod form
Inflated X
Constricted
All
inflated
2.95 to 1
Pod color
Green X Yellow
All green
882 Inflated
299
Constricted
428 Green
152 Yellow
Flower
position
Axial X Terminal
All axial
651 Axial
207 Terminal
3.14 to 1
Stem
Length
Tall X Short
All tall
787 Tall
277 Short
2.84 to 1
2.82 to 1
*Gray seed coat also had purple flowers, White seed coat had white flowers.
Selfing Dominant F2’s to produce F3 rows gave the following results:
F2 type
Mixed rows
True breeding
ratio
Round seed
372
193
1.93 to 1
Yellow cotyledon
353
166
2.13 to 1
Gray seed coat
64
36
Inflated Pod
71
29
1.78 to 1
.
2.45 to 1
Green Pod
60
40
1.50 to 1
Axial flower
67
33
2.03 to 1
Tall plant
72
28
2.57 to 1
Average ratio to heterozygote F2 to homozygote F2 was 2.06 to 1