DescriptionLab 9 – Genetics
BIOS 1610 – Summer I 2023
Emily Catania
Genetic
Termonology
Genetic Terminology
▪ Allele: a variation of a gene
▪ Homozygous: individual that has received the
same allele from both parents
▪ Heterozygous: individual that has received
different alleles from each parent
▪ The allele received can affect the phenotype
(physical appearance) of the individual
▪The actual genetic makeup of the individual is the
genotype
Genetic Terminology
▪ Some alleles (usually those that make proteins) can mask others
▪ Dominant traits are expressed
▪ Usually involve the production of a protein which causes change in the physical appearance
▪ Recessive traits can be masked by dominant traits
▪ Usually involve reduced production and lack of a protein
▪ The dominant allele will cause a physical change if present in the individual and the recessive allele
will be masked
▪ Generations:
▪ Parental (P) – these are the purebred flies that are used to initiate the experiment
▪ Both are homozygous for the trait being studied
▪ F1 – offspring of the parental generation
▪ F2 – offspring of breeding two F1 flies
Traits are linked to the chromosome on
which they reside
During meiosis, sister chromatids are
placed into separate gametes
▪ Segregation: in gametes,
each cell only receives one
of the 2 homologous
chromosomes
M1
D
R
Punnett Square
▪ Can be used to determine the odds of offspring receiving a certain
combination of alleles
▪ Write down the genotypes of each parent
▪ Write down the possible gametes each parent can make
▪ Create a Punnett Square
Performing a Cross for Autosomal Genes
▪ Capital letters indicate a trait is dominant
▪ Lowercase letters indicate a trait is
recessive
▪ We will practice using coat color: Black
(B) and white (b)
▪ Parents are heterozygous (Bb)
Performing a Cross for Autosomal Genes
▪ Phenotype: Physical appearance. The ratio is 3:1
▪ Genotype: Genetic makeup. The ratio is 1:2:1
Performing a Cross for Sex-linked Genes
▪ Sex-linked genes are found on the sex chromosomes
▪ Like people, fruit flies have X and Y chromosomes
▪ Female: XX
▪ Male: XY
Performing a Cross for Sex-linked Genes
▪ We will practice a sex-linked cross using coat color:
▪ Black (B)
▪ White (b)
▪ In cases like these, we must note which sex chromosome is
being passed on, and which allele it carries
XB
Y
XB
Xb
XBXB
XBXb
Color: Black
Color: Black
XBY
XbY
Color: Black
Color: White
Performing a Cross for Sex-linked Genes
▪ Phenotype: Physical appearance. The ratio is 2:1:1
▪ Genotype: Genetic makeup. The ratio is 1:1:1:1
XB
Y
XB
Xb
XBXB
XBXb
Color: Black
Color: Black
XBY
XbY
Color: Black
Color: White
Performing a Chi-Square
▪ Chi-Square test is a comparison of observed frequencies to expected
frequencies
▪ In a cross between two heterozygotes, the expected phenotypic ratio is 3:1
▪ Example:
▪ Out of 100 pea plants, 60 have blue flowers (B) and 40 have white flowers (b). Does this
significantly differ from the expected frequencies of a heterozygous (Bb x Bb) cross?
Performing a Chi-Square
▪ Example:
▪ Out of 100 pea plants, 60 have blue flowers (B) and 40 have white flowers (b). Does this significantly differ
from the expected frequencies of a heterozygous (Bb x Bb) cross?
1.
Determine the expected phenotypic frequency
▪ Remember, the ratio is 3:1 (75% blue : 25% white)
▪ 0.75 x 100 = 75 blue
▪ 0.25 x 100 = 25 white
2.
Determine partial chi-square values
▪ Partial chi-square value = (Observed Number – Expected Number)2/Expected Number
▪ Ex. (60-75)2/75 = 3
▪ Ex. (40-25)2/25 = 3
Performing a Chi-Square
3.
Add partial Chi-Squares and consult the
distribution table
▪3+3=6
▪ Degrees of freedom = number of categories – 1
▪ Since there are only 2 categories (blue or white), our
degrees of freedom (df) = 1. Critical value is 3.84
▪ Alpha (or p-value) of 0.05
▪ 6 > 3.84, therefore, we reject the null hypothesis
▪ This is likely not a Bb x Bb cross
df
1
2
3
4
5
alpha (=p)
0.10
2.706
4.605
6.251
7.779
9.236
0.05
3.841
5.991
7.815
9.488
11.070
0.02
5.412
7.824
9.837
11.688
13.388
0.01
6.635
9.21
11.345
13.277
15.086
0.001
10.827
13.815
16.266
18.467
20.515
6
7
8
9
10
10.645
12.017
13.362
14.684
15.987
12.592
14.067
15.507
16.919
18.307
15.033
16.622
18.168
19.679
21.161
16.812
18.475
20.090
21.666
23.209
22.457
24.322
26.125
27.877
29.588
11
12
13
14
15
17.275
18.549
19.812
21.064
22.307
19.675
21.026
22.362
23.685
24.996
22.618
24.054
25.472
26.873
28.259
24.725
26.217
27.688
29.141
30.578
31.264
32.909
34.528
36.123
37.697
16
17
18
19
20
23.542
24.769
25.989
27.204
28.412
26.296
27.587
28.869
30.144
31.41
29.633
30.995
32.346
33.687
35.02
32.000
33.409
34.805
36.191
37.566
39.252
40.790
42.312
43.82
45.315
Fruit Fly Counting Review
▪ When counting flies, take note of how many of each of the following:
▪ Red-eyed females
▪ Red-eyed males
▪ White-eyed females
▪ White-eyed males
▪ Do not only count based on eye color !!
▪ Make sure to look at both eye color and sex
▪ Write down this information in a place where you know you will be able to find/access in
future labs and for your lab report !!
Lab 9B – Meiosis and
Genetics
Meiosis and Genetics
▪ Our genetics determine who and what we are
▪ All living organisms have DNA
▪ DNA is a code that determines how any
organism functions
▪ Genes encode proteins that have physical
impacts on our appearance
Our genetics determine who and what we are
▪ Genetic material is tightly packaged in
chromosomes, which are safely stored in the
cell’s nucleus
Chromosomes are Condensed Linear DNA
▪ Condensed chromosomes are only visible during mitosis
Certain Genes can be Found on Specific
Chromosomes
▪ Karyotypes like this can only be made during cell division
Sex chromosomes
Cell Division: Mitosis
▪ Mitosis is the process of cell division
Cell Division: Mitosis
▪ Mitosis has 4 distinct phases:
Gamete Production: Meiosis
Today’s Lab
1.
Ears of Corn
▪ 3 ears of corn
▪ Each kernel is an offspring of a cross
▪ Monohybrid ear – determine approximate phenotypic ratio by counting 100 kernels and record their
phenotypes
▪ Perform Chi-Square test
▪ Dihybrid ear – construct a cross between F1 heterozygous parents for both genes
▪ Punnett Square
▪ Count 100 kernels and record phenotypes
▪ Chi-Square test
Corn Crosses
Smooth,
purple
Smooth,
yellow
Wrinkled,
purple
Wrinkled,
yellow
Today’s Lab
2. Microscope Slides
▪Whitefish mitosis
▪Plant (allium) mitosis
▪Human karyotypes
▪ Computer program hooked up to microscope to get a better view of chromosomes
▪ TA will demonstrate
Today’s Lab
3. Population Frequency of Alleles
▪ Predict which human traits are dominant or recessive
▪ Write down your data on handout and on board
▪ Copy down class data !!
▪ I will tell you which is dominant/recessive at the end of lab
Lab 9 – Mitosis and Mendelian Genetics
Name: ________________________________
1. For the monohybrid ear of corn:
a. How many kernels of each phenotype did you count? What was the approximate phenotypic
ratio? (0.5 point)
b. What are all the possible genotypes for each phenotype? (0.5 point)
c. What are the phenotypes and genotypes of the parental strains? How did you determine this? (1
point)
d. What were the chi-square test results? What do you conclude? How confident are you? Show
your work. (1.5 points)
1
2. For the dihybrid ear of corn:
a. Attach your Punnett square. What was the expected phenotypic ratio of the offspring? (2 points)
b. How many kernels of each phenotype did you count? (0.5 point)
c. What were the chi-square test results? What do you conclude? How confident are you? Show
your work. (1.5 points)
3. Examine a slide of mitosis in whitefish (an animal cell) or allium (a plant cell).
a. Find the stages of mitosis and draw images for each stage. Look at both the chromosomes and
the spindle apparatus. (2 points)
2
b. How do these stages on a slide differ from a textbook schematic? (1 point)
4. Examine the slides that show human karyotypes. Scan the slide until you find an appropriate area:
one in which the chromosomes have been replicated and the cell contents have been spread apart well
enough to observe the individual chromosomes.
a. Draw 3 replicated chromosomes from a single nucleus. On one chromosome, label the centromere
and the sister chromatids. (1 point)
b. Name 2 ways in which non-homologous chromosomes differ that allow us to identify them. (1
point)
5. Pick one trait from the population frequency study.
a. Which trait do you choose?
b. Which allele did you predict would be dominant? Why did you make this prediction? (1 point)
3
c. Was that allele prevalent in our population? (0.5 point)
d. Was your prediction correct or incorrect? Why do you think this is the case? (1 point)
4
BIOS 1610 Lab 9 – Part A
Mendelian Genetics
Objectives
1. Become familiar with the use of fruit flies for breeding experiments
2. Use a breeding experiment to determine if the allele for a phenotypic trait is dominant or recessive
3. Use a breeding experiment to determine if the locus controlling a phenotypic trait is autosomal or
sex-linked.
4. Understand the use of the chi-square test statistic to assess the goodness of fit between your observed
data and your expectations in order to reject or accept four hypotheses regarding eye color.
**Please note: This lab is a multi-week investigation so keep this handout with you for future lab
sessions in which data collection and analysis will occur.***
Introduction
Drosophila melanogaster (common fruit fly) is a diploid species, meaning it has two copies of every
gene, which are encoded on paired chromosomes. One of these chromosomes is inherited from each
parent; one from the male parent, and the other from the female parent. On each chromosome there are
genes that often differ from each other slightly. The different forms of a gene are referred to as alleles.
Each gene resides at a particular location, or locus, on each chromosome. A dominant allele is
expressed phenotypically (is outwardly observable) when present on at least one of the two parental
chromosomes. A recessive allele is masked by a dominant allele and is expressed only when paired
with another identical recessive allele. The allelic composition of a specific locus is referred to as the
genotype of an individual. When a particular locus has two copies of the same allele (one on each
chromosome), the individual with this genotype is said to be homozygous. If two different alleles are
found at a given locus, the individual is said to be heterozygous. For instance, if a dominant allele (A)
for flower color encodes for blue flowers and a recessive allele (a) encodes white flowers, various
phenotypes are possible from the three genotypes at the flower color locus. Homozygous dominant
(AA) individuals will have blue flowers. Heterozygous (Aa) individuals will also have blue flowers.
Only homozygous recessive (aa) individuals will have white flowers.
Genes are found on the various chromosomes an individual possesses. All chromosomes differ in the
number and type of genes they encode. In the case of animals, there are generally a large number of
chromosomes which encode genes involved in the growth and maintenance of an individual. However,
in animals, there are also sex chromosomes that control the expression of sex. In most animals, there is
an “X” and a “Y” chromosome. When an individual is homozygous for the X chromosome, they will
develop as females. When an individual has one X and one Y chromosome, they will develop as males.
As a result of this configuration of sex chromosomes, all offspring are either XX or XY. In this multiweek lab, you will be determining if a trait is encoded on an autosomal chromosome or a sex
chromosome.
The species of fruit fly we will use in lab today is Drosophila melanogaster. This species has been used
for genetics experiments since 1909. An enormous number of mutants have been either discovered or
induced, resulting in a variety of phenotypic traits that are easily observable and distinguishable. Some
mutations cause the loss of wings so that wingless individuals are produced. Others affect bristle (hair)
number or eye color. In today’s lab, you will gain experience studying the inheritance of eye color in
fruit flies.
Because the length of time required to follow the inheritance of traits through two generations takes
more time than we have during the shortened summer semester, we will begin our experiment with flies
that are the offspring of parents that differed in phenotype. The flies used to begin a breeding
experiment are referred to as the parental generation. The offspring of a cross between two parents are
referred to as the F1 generation. For this experiment, the parental females were white-eyed and the
male parents had red eyes.
The point of this breeding experiment is to determine which allele for eye-color (red or white) is
dominant AND whether the eye color locus is autosomal or sex-linked by following the inheritance of
eye color through successive generations.
Your Experiment
For your part of the breeding experiment, you will need to record the sex of each F1 individual and its
phenotype. You will then place one male and one female F1 individual into a new vial and let them
reproduce. These F1 individuals will mate in the new vial and the female will lay a large number of
eggs. The offspring of this F1 cross are referred to as the F2 generation. After a sufficient time has
passed (see below for the tentative schedule), you will record the sex of each F2 individual and its
phenotype. Even though not all F1s will be used for producing the F2 generation, you should count and
phenotype all flies in the F1 vials in order to determine the numbers of males and females with particular
eye colors. An accurate count of fly phenotypes is necessary to test the four hypotheses that are listed
below.
While you should be able to easily determine the eye color phenotype of an individual fly, assessing
whether it is female or male is more challenging. In the figures shown below you can notice differences
between females and males. Males are generally smaller than females although this alone is not usually
reliable. Females have a larger number of sternites and the posterior part of the abdomen is darker in
males than in females (see below).
Working with fruit flies can be tricky. In order to transfer flies from one vial to another, it is necessary
to anesthetize them (knock them out). Anesthetizing the flies will allow you time to study them, record
phenotypes and count them, and transfer them to new vials. There is a delicate balance between
anesthetizing the flies and killing them by over-anesthetizing them. Today you will use a non-toxic
substance called “fly nap”. Your TA will demonstrate the proper methodology to use. If done properly,
your anesthetized flies will wake up in a new home with a mate and will produce large numbers of F2
offspring.
Summer I schedule for breeding experiment:
The F1s need to be transferred to new vials to mate on June 1st.
The F1 parents need to be removed from the vials on June 8th.
The F2s will begin to emerge on June 15th and can be counted until June 22nd. All counting should
becompleted by June 22nd.
*Please note: once an F2 individual has been removed and counted it should not be placed back in
the same vial*
The figure below shows the different life stages of a fruit fly. You will have an opportunity to see all of
these stages as the F2 are laid as eggs, develop into larvae, and finally metamorphose into adults, either
male or female.
For this experiment you will use Punnett Squares to generate predicted ratios of genotypes and
phenotypes for the F1 and F2 generations under four hypotheses. Assume all parental individuals are
homozygous for their allelic combinations (If sex-liked, the male is called hemizygous because he only
has one X chromosome).
Hypothesis 1: Allele for white eyes is dominant and autosomal.
Predicted genotypic and associated phenotypic ratios for F1 and F2 generations:
Hypothesis 2: Allele for white eyes is recessive and autosomal.
Predicted genotypic and associated phenotypic ratios for F1 and F2 generations:
Hypothesis 3: Allele for white eyes is dominant and sex-linked.
Predicted genotypic and associated phenotypic ratios for F1 and F2 generations:
Hypothesis 4: Allele for white eyes is recessive and sex-linked.
Predicted genotypic and associated phenotypic ratios for F1 and F2 generations:
BIOS 1610: Chi-square test Statistic for Biologists
Note: Please bring a calculator to Lab 9
Objective: Become familiar with, and implement, a Chi-square test
I. Chi-square test (c2): a test of the distribution of observations into categories
The Chi-square test (abbreviated c2) is used to determine whether observations are distributed
into categories as the null hypothesis predicts they should be. This may sound conceptually obtuse, but
it’s actually one of the easiest statistical tests to understand. The most basic type of Chi-square test is a
test for goodness of fit. In a goodness of fit test, the null hypothesis directly predicts the proportion of
observations that should fall into each category.
For instance, suppose we are interested in asking whether both parents of a group of offspring
were heterozygotes (i.e. whether both parents were Aa). We would use a Chi-square goodness of fit test
to determine whether offspring genotypes are distributed into categories as this hypothesis predicts they
should be. There are three possible categories for offspring genotypes: AA, Aa, and aa. We know that,
if both parents were heterozygous, 25% of offspring should be AA, 50% should be Aa, and 25% should
be aa. We would use a c2 test to determine whether the distribution of offspring into these three
genotype categories is consistent with the distribution (1:2:1) we would expect if both parents were Aa.
Note: Almost always, the null hypothesis takes the form of “There is no difference” or “There is
no effect.” In a goodness of fit test, the null hypothesis is not “There is no difference in the number of
individuals in different categories,” since this null hypothesis would predict a 1:1:1 ratio of offspring
genotypes. Instead, in a goodness of fit test, the null hypothesis is slightly different, and can be stated
as, “There is no difference between the observed distribution and the distribution predicted by the
hypothesis under consideration.” Since we are interested in testing whether both parents were
heterozygotes, our null hypothesis in this case is “There is no difference between the observed
distribution of offspring genotypes and the distribution expected if both parents were heterozygotes.”
Suppose we observe offspring genotypes to be distributed as follows:
84 AA
122 Aa
58 aa
What distribution of offspring genotypes does the null hypothesis predict we should observe? The null
hypothesis predicts that, regardless of how many offspring there are, 25% of all offspring should be AA,
50% of all offspring should be Aa, and 25% of all offspring should be aa. Because we observed 264
offspring, in this specific case the null hypothesis predicts that we should have expected the following
numbers in each genotype category:
Genotype
Expected Number of Offspring
(if the null hypothesis were true)
AA
Aa
aa
25% of 264 = 66
50% of 264 = 132
25% of 264 = 66
NOTE: The total sum of “Expected Number of Offspring” across all categories must equal the total
number of offspring we observed (i.e. 66 + 132 + 66 must equal 264 total offspring, which it does).
This is an easy way to check for mistakes early in your calculations.
Now we can compare the numbers we actually observed in each category to the number we
expected to observe if the null hypothesis was true. This comparison is the heart and soul of the c2 test,
because the expected values are the prediction of the null hypothesis. Note that both observed and
expected values are numbers of individuals, not proportions.
NOTE: We are about to compare our actual data to the data we would expect to observe if the null
hypothesis were true. The closer these two are, the more likely the null hypothesis is true (since the null
hypothesis did a very good job of predicting what we should observe). Because the p-value expresses
the probability that the null hypothesis is true, we should calculate a high p-value if the observed and
expected values are very close. Remember that we reject the null hypothesis only if the p-value is very
low (only if it is below 0.05). Conversely, if our actual data are very different from those the null
hypothesis predicts, then it is likely that the null hypothesis is false (since the null hypothesis did a very
poor job of predicting what we should observe). We should calculate a low p-value if the observed and
expected values are very different. We should reject the null hypothesis if the p-value is less than 0.05,
since a p-value of
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