Download Corn Genetics Lab

Corn Genetics Lab
Notes From the teacher
Day 1:
Before class:
 Review the lab.
 Complete the pre-lab.
o Title and date of the lab (remember to add this lab to your table of contents).
o Purpose  1-2 sentences describing the overall goal of the lab; use complete sentences
o Pre-Lab Questions you need to number the question, rewrite the question, and then answer it
for full credit
o Lab Procedure and Data Tables 
 Do a SUBTITLE for PART A, write the procedure (numbered list) and cut out and tape
in Data Tables 2 & 3. Leave room for all the analysis questions for Part A (you will be
doing calculations here too)
 Do a SUBTITLE for PART B, write the procedure (numbered list) and cut out and tape
in Data Table 4. Leave room for all the analysis questions for Part B (you will also be
doing calculations here)
 Do a SUBTITLE for PART C, write the procedure, and cut out and tape in Data
Tables 5, 6, &7. Make sure you have room for the analysis questions.
In class:
 Complete the Lab
CORN GENETICS LAB
LEARNING OBJECTIVES
In this laboratory, you will
 Use corn to study genetic crosses
 Recognize contrasting phenotypes
 Collect data from F1 ears of corn, and analyze the results from monohybrid and dihybrid crosses
 Study dominance, segregation, and independent assortment of alleles in corn
 Compare predicted results with results obtained from actual data
At the completion of this laboratory, you should be able to
 Collect and organize data from genetic crosses
 Analyze the data using Chi-square analysis techniques
 Predict patterns of inheritance given relevant data
BACKGROUND
Mendelian Genetics and Corn
Mendelian traits refer to phenotypical features whose pattern of
inheritance follows Mendel’s theories about the inheritance of traits. Corn —
a diploid organism — has been widely used to study and illustrate Mendelian
traits. In this lab, we are going to look at two traits: kernel color and kernel
texture. In corn, the dominant gene R determines the presence of colored
aleurone. Aleurone is a protein found in maturing seeds that can give a kernel
its color. Individuals possessing one copy of the dominant gene (R) will
exhibit purple kernels. Recessive phenotypes (rr) result in yellow kernels.
Also, in corn, the dominant gene T produces the endosperm phenotype
smooth. Smooth kernels appear hard and starchy. The recessive phenotype
produces wrinkled kernels, which are shrunken in appearance. You will see
both monohybrid (single trait) and dihybrid (two traits) crosses of the color
and texture traits.
We will examine the mode of inheritance (HOW it is inherited) of
these two genes by looking at the offspring of a fertilization event between a male corn flower and a female
corn flower. Each kernel is the progeny (offspring) of one of such fertilization events. An ear of corn shows
the results of hundreds of such events. By counting the different kernel types, we should be able to determine
their pattern of inheritance.
For instance if you observed an ear of corn where 100% of the kernels were yellow what could you
deduce about the parents? Think about it. Yellow is a recessive trait so therefore the parents must be yellow.
How about an ear that exhibited 300 smooth kernels and 100 wrinkled kernels? This 3:1 ratio is indicative of a
Tt x Tt cross.
Probability and the Chi Square Analysis
Genetics, like gambling, deals with probabilities. When you flip a coin, you have the same chance of
getting a head as a tail: a one-to-one ratio. That does not mean that if you flip a coin 100 times you will always
get 50 heads and 50 tails. You might get 53 heads and 47 tails. That is probably close enough to a one-to-one
ratio that we would accept it without a second thought. But what if you got 61 heads and 39 tails? At what point
do you begin to suspect that something other than chance is at work in determining the fall of your coin?
The probabilities calculated in a Punnett Square assume that chance (and chance only) has been
operating in the assortment and recombination of alleles that gave rise to that offspring. Thus, any variation of
the observed results from the expected results are due to chance. This is known as the null hypothesis.
Therefore, if you ACCEPT your null hypothesis, you are saying that the genes are showing typical Mendelian
inheritance and any difference between the observed and expected values is simply due to chance. If you
REJECT your null hypothesis, you are saying that the genes do NOT show typical Mendelian inheritance and
therefore the genes are linked. Linked genes are genes that are located on the same chromosome and therefore
are inherited together. So, to determine if your data is due to chance, and chance alone, perform a chi square
analysis test to see if the null hypothesis is accepted or rejected. The Chi-square test, x2, is a statistical test used
to determine how well observed ratios fit expected ratios. The x2 value consists of the summation of these
values for all phenotypes.
X2 =
(Observed – Expected)2
Expected
Table 1: Chi-Square Values and Probabilities
Probability
Degrees of Freedom (df)
1
2
3
4
5
0.05
3.84
5.99
7.82
9.49
11.1
0.01
6.64
9.21
11.3
13.2
15.1
0.001
10.8
13.8
16.3
18.5
20.5
Remember that degrees of freedom (df) is n-1. Monohybrid cross df = 1
Dihybrid cross df = 3
--------------------------------------------------------------------------------------------------------------------------------------PRELAB QUESTIONS:
1. What organism will we be using to study genetics in this activity?
2. What traits will we be examining?
3. What is the difference between a monohybrid and dihybrid cross?
4. Two F1 parents are crossed: Rr x rr. Predict the offspring. (Do a Punnett Square)
5. The following offspring are recorded: 52 smooth, 48 wrinkled. Determine the parental genotypes.
6. How can you test if your data is due to the chance assortment of alleles during meiosis?
7. You collected the following data: 433 kernels, including 334 purple and 99 yellow. This ear of corn
was produced by monohybrid cross (Rr x Rr). Do your expected results match your observed results?
Calculate a chi square value and determine the probability. (df = 1)
8. How can you tell if genes are linked or not?
9. Why are Drosophila often used in labs for genetics experiments?
10. What is a “wild type” fly?
--------------------------------------------------------------------------------------------------------------------------------------PART A: Analysis of Ear A & B
MATERIALS  PART A:
Corn Ear A
Corn Ear B
PROCEDURE  PART A:
1. Obtain an ear of corn labeled A.
2. Determine whether it is a monohybrid (one trait) or dihybrid (two traits) cross.
3. In Table 2, record one phenotype in each column.
4. Count 10 rows of kernels and record how many kernels express each phenotype.
5. Put your data into the Class Data Chart on Mrs. Willis’ computer.
6. Once all groups have recorded their data, fill in the class data portion of Table 2.
Table 2. Corn Ear A Data
Number of kernels
with phenotype:
Number of kernels
with phenotype:
Total
number of
kernels:
Your Data
(OBSRVED)
Class Data
(OBSERVED)
Class Data
(EXPECTED)
7. Obtain an ear of corn labeled B.
8. Repeat steps #2-6 above and record your data in Table 3.
Table 3. Corn Ear B Data
Number of kernels
with phenotype:
Number of kernels
with phenotype:
Total
number of
kernels:
Your Data
(OBSRVED)
Class Data
(OBSERVED)
Class Data
(EXPECTED)
ANALYSIS QUESTIONS  PART A:
Use THE CLASS DATA for your analysis!
1. Calculate the percent of each phenotype present on Ear A.
2. Determine the parental phenotypes and genotypes based on the data for Ear A.
3. Make and solve a Punnett Square for the parental cross to determine the expected offspring for Ear A.
4. Based on your total number of kernels for Ear A, and the expected ratios you found in your Punnett
square, what are you expected numbers for each phenotype on Ear A? Fill these numbers in on Table 2.
5. Using your expected numbers you calculated in #4, and the observed from your data table, do a chi
square test to determine if the observed results match the expected results for Ear A. What is your chi
square value? Show your work in your lab notebook.
6. What is the critical value of p for this chi square test?
7. What is your null hypothesis for Part A, Ear A (this is also your null for Ear B)?
8. Based on your chi square value, would you accept or reject the null hypothesis?
9. Calculate the percent of each phenotype present on Ear B.
10. Determine the parental phenotypes and genotypes based on the data for Ear B.
11. Make and solve a Punnett Square for the parental cross to determine the expected offspring for Ear B.
12. Based on your total number of kernels for Ear B, and the expected ratios you found in your Punnett
square, what are you expected numbers for each phenotype on Ear B? Fill these numbers in on Table 3.
13. Using your expected numbers you calculated in #12, and the observed from your data table, do a chi
square test to determine if the observed results match the expected results for Ear B. What is your chi
square value? Show your work in your lab notebook.
14. What is the critical value of p for this chi square test?
15. Based on your chi square value, would you accept or reject the null hypothesis?
--------------------------------------------------------------------------------------------------------------------------------------PART B: Analysis of Ear C
MATERIALS  PART B:
Corn Ear C
PROCEDURE  PART B:
1. Obtain an ear of corn labeled C. This corn was created by crossing two individuals heterozygous for
both traits.
2. Determine whether it is a monohybrid (one trait) or dihybrid (two traits) cross.
3. In Table 4, record one phenotype in each column. (You will have 4 phenotypes here)
4. Count ALL of the kernels and record how many kernels express each phenotype (you can ignore the
ones under the sticker that you cannot see). Fill in the OBSERVED DATA in Table 4.
Table 4. Corn Ear C Data
Number of
kernels with
phenotype:
Number of
kernels with
phenotype:
Number of
kernels with
phenotype:
Number of
kernels with
phenotype:
Total
number
of
kernels:
Your data
OBSERVED
DATA
EXPECTED
DATA
ANALYSIS QUESTIONS  PART B:
Use YOUR DATA for your analysis!
1. Make and solve a Punnett Square for the parental cross to determine the expected offspring. HINT:
look back at Step 1 for the parents genotypes.
2. Using your total number of kernels and the expected ratios you calculated in #1, what are the expected
values for each of the 4 phenotypes? Record these values in Table 4 under EXPECTED DATA.
3. Using your expected values from #2 and your observed values from Table 4, perform a chi square
analysis test. Make sure to show all work in your lab notebook. What is your chi square value?
4. What is your null hypothesis for Part B, Ear C, of this lab?
5. What is your p value for a dihybrid cross (df = 3)?
6. Would you accept or reject your null hypothesis and why?
7. Are the color and texture genes linked?
--------------------------------------------------------------------------------------------------------------------------------------PART C: Analysis of Drosophila
Drosophila melanogaster, the common fruit fly, is a model organism for genetic studies. You will often be
asked questions dealing with this organism. The reason it is so widely used is because it is easily cultured in
the lab, has a short generation time, can produce many offspring, and has traits that can easily be seen. In fruit
fly genetics, the normal fly is called a "wild type" and any fly exhibiting a phenotypic mutation is called a
"mutant". Mutant flies are given names that generally denote the type of mutation the fly exhibits. For example,
the mutant "ebony" has a much darker body than the wild type fly.


Mutant (w) – ebony body color
Wild Type (W) – normal body color
PROCEDURE  PART C:
1. Answer all the ANALYSIS QUESTIONS for this section of the lab. Make sure to show all calculations.
BACKGROUND INFORMATION  PART C:
Assume you identify 100 flies and record the following data for the offspring of an unknown cross involving a
single trait. Body color in Drosophila is found on chromosome 3 (it is not x-linked  so you do not have to
take into account males vs. females and can just use the TOTAL numbers to do your chi square. If it WAS xlinked, we would expect very different numbers between males and female flies).
Table 5. Fruit Fly Data – Unknown Cross
MALE
FEMALE
TOTAL
EXPECTED
OBSERVED
Wild Type
40
33
73
Mutant
15
12
27
ANALYSIS QUESTIONS  PART C:
1. Assuming the parents are both heterozygous for body color, make and solve a Punnett Square for the
parental cross to determine the expected results.
2. Using the totals from Table 5 and the expected ratios from #1, what are the expected numbers for each
phenotype? Record those figures in Table 5.
3. Complete a chi square analysis to test the null hypothesis. Make sure to show all your work. What is
your chi square value? Do you accept or reject the null?
You identify 532 flies and record the following data for the offspring of a cross involving two traits. Parent one
was heterozygous for both traits. Parent two was mutant for both traits.




Mutant (w) – ebony body color
Wild Type (W) – normal body color
Mutant (a) – short wings
Wild Type (A) – long wings
Table 6. Fruit Fly Data – Dihybrid Heterozygote crossed with Dihybrid Mutant
OBSERVED
DATA
EXPECTED
DATA
Wild type
color and
mutant wings
Mutant color
and wild type
wings
Wild type
for both
traits
87
101
298
Mutant for
TOTAL
both traits
46
4. Determine the genotypes for both parents.
5. Make and solve a Punnett Square for the parental cross to determine the expected results. Find the ratio
and then determine the expected values for each genotype. Fill in Table 6.
6. Use the chi square test to determine if color and wings are linked genes. Show all your work and make
sure to include your chi square value, whether you accepted or rejected the null, and if the genes are
linked or not.
7. The percent recombination between two genes indicates the map distance between them.
1% recombination = 1 map unit (m.u.). To determine the map distance between a pair of loci, complete
Table 7:
Table 7. Calculating Map Units Separating Body Color and Wing Length Genes
Answer the Question or SHOW YOUR
WORK Here
Identify the PARENTAL
PHENOTYPES: (What is the
phenotype of each parent? There
will be 2)
Identify the RECOMBINANT
PHENOTYPES: (What are the
other 2 phenotypes that are not
expressed by either parent? There
will be 2)
Count the # of recombination
events (# of offspring that show a
recombinant phenotype)
Calculate the recombination
frequency:
(# recombinants/ total # of
offspring) x 100
So, how many map units separate
the body color gene and the wing
length gene on the chromosomes of
Fruit Flies?