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Title: BTEC Applied Science Unit 18 Assignment 3
Description: Unit 18: Genetics and Genetic Engineering – Assignment 3 Inheritance Task 1 Complete the worksheet on inheritance of one (monohybrid) or two (dihybrid)characteristics. Follow the instructions to carry out a series of genetic crosses to study the inheritance of 2 characteristics in Drosophila, from the parental to F2 generation. You must provide a method and a result table showing the number of flies observed for each phenotype in the F2 generation. This provides evidence for P5 Task 2 You must use your knowledge of inheritance, demonstrated in task 1, to predict the outcomes of your Drosophila genetic crosses. You must calculate the expected number of flies for each phenotype, using the expected phenotypes ratio. Complete all the worksheets in the case studies on inheritance in humans. This provides evidence for M3 Task 3 Carry out a Chi2 test to compare the observed and expected ratios from one of the genetic crosses that you have performed with Drosophila. (also gives Unit 8, P2) Interpret the results of the Chi2 test by explaining whether or not the null hypothesis should be accepted. (also gives Unit 8, M2) Evaluate the extent to which you think that your result and interpretation of the Chi2 test are valid. (also gives Unit 8, D2) This provides evidence for D2 (also provides evidence for Unit 8 P2,M2,D2) Exam board is Pearson ALL ASSIGNMENTS I HAVE UPLOADED ARE DISTINCTION GRADED.

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Unit 18 – Assignment 3 – P5
Worksheets:

Fruit fly experiment

Null hypothesis = There is no significant difference between the observed the expected results
...

5 x male ebony body, vestigial wing flies
...

diethyl ether (ethoxyethane), cotton buds
...


Method – Day 1:
1
...

2
...

3
...

4
...

5
...

6
...

7
...


Method - Day 5:
1
...

2
...

3
...
These are the F1 generation
...
Inspect them, count them, and note what type and sex each one is
...
Do this until day 15
...


Method - Day 20:
Eggs should be laid and larvae visible by now
...


Method - Day 20 and onwards:
1
...

2
...


Results table:
Alleles:
G – Grey
g - Black
N – Long wings
n - Short wings

F1 parents genotype: ggnn x GGNN
(They are both homozygous)

Parents gametes = GN, Gn, gN, gn

Punnet square:
gn
gn
gn
gn

GN
GgNn
GgNn
GgNn
GgNn

GN
GgNn
GgNn
GgNn
GgNn

GN
GgNn
GgNn
GgNn
GgNn

GN
GgNn
GgNn
GgNn
GgNn

Phenotype is GgNn
...


Unit 18 – Assignment 3 – M3
Case Studies:

Fruit fly experiment
Alleles:
G – Grey
g - Black
N – Long wings
n - Short wings

F1 parents genotype: GgNn x GgNn

Parents gametes = GN, Gn, gN, gn

Punnet square:

GN
Gn
gN
gn

GN
GGNN
GGNn
GgNN
GgNn

Gn
GGNn
GGnn
GgNn
Ggnn

gN
GgNN
GgNn
ggNN
ggNn

gn
GgNn
Ggnn
ggNn
ggnn

F2 phenotypes:










GGNN
GGNn
GgNN
GgNn
GgNN
GgNn
Ggnn
GGnn
Ggnn

Grey long = 9
Grey short = 3
Black normal = 3
Black short = 1

Unit 18 – Assignment 3 – D2
Long wings and grey body = 64
Long wings and black body = 25
Short wings and grey body = 22
Short wings and black body = 5
Total flies = 116

Expected ratios = 9:3:3:1

Expected number of flies:
Long wings and grey body = (116/16) x 9 = 65
Long wings and black body = (116/16) x 3 = 22

Short wings and grey body = (116/16) x 3 = 22
Short wings and black body = (116/16) x 1 = 7

Observed Expected
Long wings and grey body
Long wings and black body
Short wings and grey body
Short wings and black body

64
25
22
5

65
22
22
7

(OE)
-1
3
0
-2

(O-E)2 (O-E)2/E
1
9
0
4

0
...
41
0
0
...
995

P value = 0
...
com/slide/256718/

When I compare my p value to the table, I identified there to be 3 degrees of freedom
...
82 or higher for the hypothesis to be rejected, and my P value was 0
...
The bigger the sample is, the
more valid it would be
...
This is due to many errors that could have occurred during the
final observation of the fruit flies
...
Also, if someone didn’t have very good
eyesight, or was even colour blind, then it would have been even more difficult
...
This error would affect
the calculations and the reliability and the accuracy of the explanations
...
If the test tube still has maggots,
then we would not have nee able to get the full number of offspring counted
...
This means
tha ti would have rejected the null hypothesis
...
Therefore it represents all the data
...
It can also be used to see if there is a difference between
the many groups of participants
...


Weaknesses:





It is based on a theory of probability and is subject to or involving chance variation
...

All participants in the experiment must be independent
...
If this were so, the chi squared analysis would not be appropriate
...

When carrying out chi squared, so that participants being identified must be randomly
selected from the total population
...



Title: BTEC Applied Science Unit 18 Assignment 3
Description: Unit 18: Genetics and Genetic Engineering – Assignment 3 Inheritance Task 1 Complete the worksheet on inheritance of one (monohybrid) or two (dihybrid)characteristics. Follow the instructions to carry out a series of genetic crosses to study the inheritance of 2 characteristics in Drosophila, from the parental to F2 generation. You must provide a method and a result table showing the number of flies observed for each phenotype in the F2 generation. This provides evidence for P5 Task 2 You must use your knowledge of inheritance, demonstrated in task 1, to predict the outcomes of your Drosophila genetic crosses. You must calculate the expected number of flies for each phenotype, using the expected phenotypes ratio. Complete all the worksheets in the case studies on inheritance in humans. This provides evidence for M3 Task 3 Carry out a Chi2 test to compare the observed and expected ratios from one of the genetic crosses that you have performed with Drosophila. (also gives Unit 8, P2) Interpret the results of the Chi2 test by explaining whether or not the null hypothesis should be accepted. (also gives Unit 8, M2) Evaluate the extent to which you think that your result and interpretation of the Chi2 test are valid. (also gives Unit 8, D2) This provides evidence for D2 (also provides evidence for Unit 8 P2,M2,D2) Exam board is Pearson ALL ASSIGNMENTS I HAVE UPLOADED ARE DISTINCTION GRADED.