Genetics Assignment: Drosophila Cross Analysis and Gene Mapping

This assignment delves into Drosophila genetics, presenting solutions for analyzing crosses and mapping genes. Part A focuses on a cross between two Drosophila strains, employing genetic symbols and hypotheses to determine phenotypic and genotypic ratios, using a breeding table and chi-squ...Read More

Running head: GENETICS
1
Genetics
Name of the student
Name of university

GENETICS 2
Genetics
Part A
a) Data set for a cross between two breeding Drosophila strains that differ in traits
Wild type, female 1586
Wild type, males 807
Ruby eyes, males 770
Curved wings, females 517
Ruby eyes, curved wings, males 251
Curved wings, males 265
b) Genetic symbols
Wild type- recessive +/+
Ruby eyes- X-linked Female X+ X+, X+ XW; Male X+ Y
Curved wings- dominant- CC, Cc
c) Genetic hypothesis
The allele for curved wings is dominant while and the allele for ruby eyes is x linked and
dominant
Curved wings are dominant. This means that the gene will be expressed in both homozygous
state and heterozygous state.
P – CC(male) x Cc (female)
F1- CC, Cc, CC, Cc
F2- CC, Cc, Cc, cc
The F1 generation will all have curved wings. However, they will different genotypes. Some are
heterozygous while others are homozygous. The genotypic ration will be 1:1. The F2 generation
will have a larger propablty of having curved wings as compared to the probability of not having
curved wings. The phenotypic ration will be 3:1 and the genotypic ratio will be 1:2:1

GENETICS 3
Ruby eyes are sex linked genes. They are found of the X chromosomes. Females have two X
chromosomes while males have one X chromosome. The gene is dominant. It is expressed even
in homozygous state.
P- X+ X+ (female), XW Y (male)
F1- X+ XW, X+ Y, X+ XW , X+ Y
F2- X+ X+, X+ Y, X+ XW, XW Y
The F1 generation will consist of all offspring with red eyes. This is because they all posses an X
chromosomes. The genotypic ration for the F1 generation is 2:2. The females have the same
genotype while the males have the same genotype. The F2 generation will consist of females
with ruby eyes, one male with ruby eyes and one male with white eyes. The two female have the
same phenotype but different genotype. The genotypic ratio for F2 generation is 1:1:1:1
d) Breeding table to analyse the set hypothesis
X+ X X+ Y X+ XW XW Y
X+ X+, 1586 807 770 517
X+ Y 770 770 517
X+ XW 1586 517 251 265
XW Y 517 251 265
F1 generation phenotypic ratio = 1
Genotypic ration = 2:2
F2 generation phenotypic ratio = 3:1
Genotypic ration = 1:1:1:1
e) Chi square
The chi square helps to analyze the result got form the practical and the theoretical results. These
variation between is very useful in understanding genetics. The variation occurs as a result of
mutation in the sequences, evolution of genes and death of some organisms.
X2 = ∑ (O-E)2
E
= (770-251)2
251
= 1073.151394422311

GENETICS 4
Part B
a) Data set of the results of a test cross using male flies that are homozygous recessive
for three traits and females that are heterozygous for the same three traits
+++ 139
abc 140
++c 154
a+c 158
ab+ 160
+b+ 155
a++ 225
+bc 218
b) Indicating Parental class, Double Crossover (DCO) class and Single Crossover
(SCO) classes
Abc 1 parental class
+++ 139 DCO class
abc 140 DCO class
++c 154 NCO class
a+c 158 SCO (1) class
ab+ 160 NCO class
+b+ 155 SCO (1) class
a++ 225 SCO (2) class
+bc 218 SCO (2) class
c) Finding the Gene Order
No crossover (NCO) for this test cross is:

GENETICS 5
++z and ab+
Double Crossover (DCO) for this test cross is:
+++ and abc
This means that c is at the center making the order of genes bca
d) Rewriting the table using correct gene order
++c 154 NCO class
ab+ 160 NCO class
a+c 158 SCO (1) class
+b+ 155 SCO (1) class
a++ 225 SCO (2) class
+bc 218 SCO (2) class
+++ 139 DCO class
abc 140 DCO class
SCO (1) and SCO(1)are compatible while SCO (2) and SCO (2) are compatible
e) Map distances (Recombination frequencies) for the genes in the test cross
Map distances helps to identify the exact position of a gene in the chromosome. Through the
map distance the order of genes can also be identified and mutations can be detected early.
To determine distance between genes = SCO +DCO x 100
Total number of offspring
Therefore a to c distance
=DCO between a and c= 1+3=4
=SCO between a and c=158+155=313
= DCO +SCO= 4+313=317

GENETICS 6
= 317/1379*100
=22.99
C to b distance
=DCO between c and b= 1+3=4
=SCO between c and b=318+255=573
= DCO +SCO= 4+573=577
= 577/1379*100
=41.84
f) Genetic map indicating the map distances
. .
41.84mu 22.9mu
b c a

GENETICS 7
References
Combs, P. A., & Eisen, M. B. (2017). Genome-wide measurement of spatial expression in
patterning mutants of Drosophila melanogaster. F1000Research, 6.
Hunter, C. M., Huang, W., Mackay, T. F., & Singh, N. D. (2016). The genetic architecture of
natural variation in recombination rate in Drosophila melanogaster. PLoS genetics, 12(4),
e1005951.