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expo8labreport
Hi VGR,Please see attached.
This is the edited version for the lab report of expo 8. I just deleted Pages 7 to 8. They are from expo 7. What I usually do is I used the previous lab report for the format and just erase the content and replace a new one. In this case I missed deleting that portion. My sincere apologiesAnd thank you so much for reviewing.SincerelyRoel
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, hitc
hh
iker’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 o
ff
spring if parents are heterozygous brown-eyed individuals with dimpled chin.
Data Tables:
Su
mm
ary Table:
|
Trait |
Phenotype | Genotype | ||||||||
|
1.Dimpled chin |
recessive |
dd |
||||||||
|
2.Free ear lobes |
ff | |||||||||
|
3.Ability to taste PTC |
pp |
|||||||||
|
4.Interlocking fingers |
||||||||||
|
5.Mid-digital hair |
dominant |
MM(homozygous dominant) or Mm(heterozygous dominant) |
||||||||
|
6.Bent little finger |
bb |
|||||||||
|
7.Widow’s peak |
ww |
|||||||||
|
8. Hitchhiker’s thumb |
hh | |||||||||
|
9. Pigmented irises |
ii |
|||||||||
|
10. Long palmar muscle |
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 B
bD
d.
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 4×4 Punnet
Square:
Punnet Square:
|
|
BD | Bd | bD | bd | |||||||||||||||||||||||||
|
BBDD |
BBDd |
BbDD |
BbDd |
||||||||||||||||||||||||||
|
BBdd |
Bbdd |
||||||||||||||||||||||||||||
|
bbDD |
bbDd |
||||||||||||||||||||||||||||
|
bbdd |
Phenotype Classification:
|
BD |
|
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-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
BBdd
BbDd
Bbdd
bD
BbDD
BbDd
bbDD
bbDd
bd
BbDd
Bbdd
bbDd
bbdd
|
orange |
|
| blue | |
|
light blue |
|
|
dark blue |
|
| green | |
|
white colors |
BBDD,BBdd,bbDD,bbdd |
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 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.
