Student's Name: Date of Experiment: September 21, 2013 Date Report Submitted: September 28, 2013 Title: Experiment 8: Phenotype and Genotype Purpose: The purpose of the experiment is to let students compare, analyze and determine phenotype and genotype Procedure: The student’s phenotype and genotype were determined for dimpled chin, free ear lobe, ability to taste PTC, interlocking fingers, mid-digital hair, bent little finger, Widow’s peak, hitchhiker’s thumb, pigmented irises and long palmar muscle in accordance with the conditions for having a dominant and recessive trait. An exercise was done to determine genotype and phenotype ratios possible for offspring if parents are heterozygous brown-eyed individuals with dimpled chin. Data Tables: Summary Table: Trait Phenotype Genotype 1.Dimpled chin recessive dd 2.Free ear lobes recessive ff 3.Ability to taste PTC recessive pp 4.Interlocking fingers recessive ff 5.Mid-digital hair dominant MM(homozygous dominant) or Mm(heterozygous dominant) 6.Bent little finger recessive bb 7.Widow’s peak recessive ww 8. Hitchhiker’s thumb recessive hh 9. Pigmented irises recessive ii 10. Long palmar muscle recessive mm Observations: From the table, it can be deduced that the student is mostly of recessive type for the traits specified. It was only the mid-digital hair that was the dominant trait. It can be deduced further that a phenotype that was recessive, the genotype was a homozygous recessive. For a dominant phenotype, the genotype was either a homozygous or heterozygous dominant. Questions/Exercise: Refer to the previous experiment and construct a Punnett square showing both the genotype and phenotype ratios possible if two heterozygous brown-eyed individuals with dimpled chins were to have children. Your Punnett square will be 4 x 4squares. Assume both independent assortment and segregation are occurring. Solution : Since both parents are heterozygous and there are two genes, each parent has the following genotype BbDd. B = brown eyes b = other eye color (non-brown) D= with dimple chin d= not dimple chin There are four combinations possible BD, Bd,bD and bd then constructing the 4x4 Punnet Square: Punnet Square: BD Bd bD bd BD BBDD BBDd BbDD BbDd Bd BBDd BBdd BbDd Bbdd bD BbDD BbDd bbDD bbDd bd BbDd Bbdd bbDd bbdd Phenotype Classification: BD Bd bD bd BD BBDD BBDd BbDD BbDd Bd BBDd BBdd BbDd Bbdd bD BbDD BbDd bbDD bbDd bd BbDd Bbdd bbDd bbdd legend: green = offspring with brown eyes and with dimple chins yellow= offspring with brown eyes but no dimple chins blue = offspring with non-brown eyes color but with dimple chins red = offspring with non-brown eyes color and no dimple chins Interpretation: From the Punnet square and from the color scheme, it can be seen that there are 9 offspring with brown eyes and with dimple chins, 3 offspring with brown eyes but no dimple chins, 3 offspring with non-bro.

1. Student's Name: Date of
Experiment: September 21, 2013
Date Report Submitted: September 28, 2013
Title: Experiment 8: Phenotype and Genotype
Purpose:
The purpose of the experiment is to let students compare,
analyze and determine
phenotype and genotype
Procedure:
The student’s phenotype and genotype were determined for
dimpled chin, free ear lobe, ability to taste PTC, interlocking
fingers, mid-digital hair, bent little finger, Widow’s peak,
hitchhiker’s thumb, pigmented irises and long palmar muscle in
accordance with the conditions for having a dominant and
recessive trait. An exercise was done to determine genotype and
phenotype ratios possible for offspring if parents are
heterozygous brown-eyed individuals with dimpled chin.
Data Tables:
Summary Table:
Trait
Phenotype
Genotype
1.Dimpled chin
recessive
dd
2.Free ear lobes
recessive
ff
3.Ability to taste PTC
recessive
pp
4.Interlocking fingers

2. recessive
ff
5.Mid-digital hair
dominant
MM(homozygous dominant) or Mm(heterozygous dominant)
6.Bent little finger
recessive
bb
7.Widow’s peak
recessive
ww
8. Hitchhiker’s thumb
recessive
hh
9. Pigmented irises
recessive
ii
10. Long palmar muscle
recessive
mm
Observations:
From the table, it can be deduced that the student is mostly of
recessive type for the traits specified. It was only the mid-
digital hair that was the dominant trait. It can be deduced
further that a phenotype that was recessive, the genotype was a
homozygous recessive. For a dominant phenotype, the genotype
was either a homozygous or heterozygous dominant.
Questions/Exercise:
Refer to the previous experiment and construct a Punnett square
showing both the genotype and phenotype ratios possible if two
heterozygous brown-eyed individuals with dimpled chins were

3. to have children. Your Punnett square will be 4 x 4squares.
Assume both independent assortment and segregation are
occurring.
Solution
:
Since both parents are heterozygous and there are two genes,
each parent has the following genotype BbDd.
B = brown eyes
b = other eye color (non-brown)
D= with dimple chin
d= not dimple chin
There are four combinations possible BD, Bd,bD and bd then
constructing the 4x4 Punnet
Square:
Punnet Square:

4. BD
Bd
bD
bd
BD
BBDD
BBDd
BbDD
BbDd
Bd
BBDd
BBdd
BbDd
Bbdd
bD
BbDD
BbDd
bbDD
bbDd
bd
BbDd
Bbdd
bbDd
bbdd

5. Phenotype Classification:
BD
Bd
bD
bd
BD
BBDD
BBDd
BbDD
BbDd
Bd
BBDd
BBdd
BbDd
Bbdd
bD
BbDD
BbDd
bbDD
bbDd
bd
BbDd
Bbdd

6. bbDd
bbdd
legend:
green =
offspring with brown eyes and with dimple chins
yellow=
offspring with brown eyes but no dimple chins
blue =
offspring with non-brown eyes color but with dimple chins
red =

7. offspring with non-brown eyes color and no dimple chins
Interpretation:
From the Punnet square and from the color scheme, it can be
seen that there are 9 offspring with brown eyes and with dimple
chins, 3 offspring with brown eyes but no dimple chins, 3
offspring with non-brown eyes color but with dimple chins and
1 offspring with non-brown eyes color and no dimple chins.
Phenotype ratio: 9:3:3:1
Genotype Classification:
BD
Bd
bD
bd
BD
BBDD
BBDd
BbDD
BbDd
Bd
BBDd

8. BBdd
BbDd
Bbdd
bD
BbDD
BbDd
bbDD
bbDd
bd
BbDd
Bbdd
bbDd
bbdd
orange
BBDd
blue
BbDD
light blue
Bbdd
dark blue
bbDd
green
BbDd
white colors
BBDD,BBdd,bbDD,bbdd

9. Interpretation:
From the Punnet Square and color scheme, it can be seen that
there are 1 BBDd, 2 BbDD, 2 Bbdd,2 bbDd, 4 BbDd, and 1
BBDD,BBdd,bbDD,bbdd.
Genotype ratio: 1:2:2:2:4:1:1:1:1
Conclusion:
Phenotype is the physical trait of an individual. Since its
physical, it can be determined from the outward appearance of a
person. An individual can be a dominant or recessive
phenotype. This was emphasized in the experiment as student
was either dominant or recessive for the traits specified.
Genotype can be considered an inside trait of an individual and
thus cannot not be determined immediately. However, it can be
determined by knowing the phenotype of a person. Just like
phenotype, genotype can be dominant or recessive. If a person
is a dominant phenotype, the person’s genotype can be
homozygous or heterozygous. In the experiment, the student
was dominant phenotype for mid-digital finger so the genotype
can be represented as MM (homozygous) or Mm (heterozygous)
where capital letter M represents dominant and small letter m

10. represents recessive. If a person is a recessive phenotype, the
person’s genotype is homozygous recessive. For most of the
traits specified in the experiment, the student was recessive
phenotype and therefore genotype was mostly homozygous
recessive.
The exercise provided in the experiment was an excellent
demonstration on how Mendel’s law of segregation affects the
outcome of the characteristics of offspring. Although parents
both have brown eyes and with dimple chins, offspring may
have different characteristics with ratio of 9:3:3:1 for
phenotype. Therefore the couple with both brown eyes and with
dimple chins may even produce a child that doesn’t have a
dimple chin and brown eyes at all. The separation of alleles
during gamete formation is responsible for these different
characteristics.
DataIDSalCompaMidAgeEESSRGRaiseDegGen1Gr1581.017573
485805.70METhe ongoing question that the weekly assignments
will focus on is: Are males and females paid the same for equal
work (under the Equal Pay
Act)?2270.870315280703.90MBNote: to simplfy the analysis,
we will assume that jobs within each grade comprise equal
work.3341.096313075513.61FB4661.15757421001605.51METh
e column labels in the table
mean:5470.9794836901605.71MDID – Employee sample

11. numberSal – Salary in thousands6761.1346736701204.51MFAge
– Age in yearsEES – Appraisal rating (Employee evaluation
score)7411.0254032100815.71FCSER – Years of serviceG –
Gender (0 = male, 1 = female)8231.000233290915.81FAMid –
salary grade midpointRaise – percent of last
raise9771.149674910010041MFGrade – job/pay gradeDeg (0=
BSBA 1 = MS)10220.956233080714.71FAGen1 (Male or
Female)Compa - salary divided by
midpoint11231.00023411001914.81FA12601.0525752952204.50
ME13421.0504030100214.70FC14241.04323329012161FA1524
1.043233280814.91FA16471.175404490405.70MC17691.21057
27553131FE18361.1613131801115.60FB19241.043233285104.6
1MA20341.0963144701614.80FB21761.1346743951306.31MF2
2571.187484865613.81FD23231.000233665613.30FA24501.041
483075913.80FD25241.0432341704040MA26241.04323229521
6.20FA27401.000403580703.91MC28751.119674495914.40FF2
9721.074675295505.40MF30491.0204845901804.30MD31241.0
43232960413.91FA32280.903312595405.60MB33641.12257359
0905.51ME34280.903312680204.91MB35241.043232390415.30
FA36231.000232775314.30FA37220.956232295216.20FA38560
.9825745951104.50ME39351.129312790615.50FB40251.086232
490206.30MA41431.075402580504.30MC42241.043233210081
5.71FA43771.1496742952015.50FF44601.0525745901605.21M
E45551.145483695815.21FD46651.1405739752003.91ME47621
.087573795505.51ME48651.1405734901115.31FE49601.052574

12. 1952106.60ME50661.1575738801204.60ME
Week 1Week 1.Describing the data.1. Using the Excel Analysis
ToolPak function descriptive statistics, generate descriptive
statistics for the salary data.Which variables does this function
not work properly for, even though we have some excel
generated results?2. Sort the data by Gen or Gen 1 (into males
and females) and find the mean and standard deviation for each
gender for the following variables:sal, compa, age, sr and
raise.Use the descriptive stats function for one gender and the
Fx functions (average and stdev) for the other.3. What is the
probability distribution table for a:a. Randomly selected
person being a male in a specific grade?b. Randomly
selected person being in a specific grade?4. Find:a. The z score
for each male salary, based on only the male salaries.b. The z
score for each female salary, based on only the female
salaries.5. Repeat question 4 for compa for each gender.6.
What conclusions can you make about the issue of male and
female pay equality? Are all of the results consistent? If not,
why not?
Week 2 Week 2Testing means1Is either the male or female
salary equal to the overall mean salary?(Two hypotheses tests -
1 sample tests)2Are the male and female salaries statistically
equal to each other?3Are the male and female compas equal to
each other?4. If the salary and compa mean tests in questions 3
and 4 provide different equality results,which would be more

13. appropriate to use in answering the question about salary
equity? Why?5. What other information would you like to
know to answer the question about salary equity between the
genders? Why?
Week 3Week 31. Is the average salary the same for each of
the grade levels? (Assume equal variance, and use the analysis
toolpak function ANOVA.)Set up the input table/range to use as
follows: Put all of the salary values for each grade under the
appropriate grade label.ABCDEF2. The factorial ANOVA
with only 2 variables can be done with the Analysis ToolPak
function 2-Way ANOVA with replication. Set up a data input
table like the following:GradeGenderABCDEFMFFor each
empty cell randomly pick a male or female salary from each
grade.Interpret the results. Are the average salaries for each
gender (listed as sample) equal?Are the average salaries for
each grade (listed as column) equal?3. Repeat question 2 for
the compa values.GradeGenderABCDEFMFFor each empty cell
randomly pick a male or female salary from each grade.Interpret
the results. Are the average compas for each gender (listed as
sample) equal?Are the average compas for each grade (listed as
column) equal?4. Pick any other variable you are interested in
and do a simple 2-way ANOVA without replication. Why did
you pick this variable and what do the results show?5. What
are your conclusions about salary equity now?
Week 4Week 4Confidence Intervals and Chi Square (CHs 11 -

14. 12)Q1Q2Let's look at some other factors that might influence
pay.GrDegGen1SalA0F341. Is the probability of having a
graduate degree independent of the grade the employee is
in?A0F41C0F772. Construct a 95% confidence interval on
the mean service for each gender? Do they
intersect?C0F55D1M773. Are males and females distributed
across grades in a similar pattern?D1M604. Do 95%
confidence intervals on the mean length of service for each
gender intersect?5. How do you interpret these results in
light of our equity question?
Week 5Week 5 Correlation and Regression1. Create a
correlation table for the variables in our data set. (Use analysis
ToolPak function Correlation.)2. Create a multiple regression
equation (using the Analysis ToolPak function Regression) to
predict either salary or compa using the mid(a substitute
variable for grade level), age, ees, sr, raise, and deg variables.
(Note: since salary and compa are different ways ofexpressing
an employee’s salary, we do not want to have both used in the
same regression.)3. Based on all of your results to date, is
gender a factor in the pay practices of this company? Why or
why not?4. In looking at equal pay issues across an entire
company, which is a better variable to use – compa or salary?
Why?5. Why did the single factor tests and analysis (such as
t and single factor ANOVA tests on salary equality) not provide
a complete answer to our salary equality question?What

15. outcomes in your life or work might benefit from a multiple
regression examination rather than a simpler one varable test?