Download Chromosome Mapping Lab

CHAPTER 17
HANDOUT
BLM 17.2.3
Thought Lab 17.1: Mapping
Chromosomes
Purpose: To construct a chromosome map.
Steps to determine map distance
Step
Example
Perform a cross between a fly that is known to
be heterozygous for both traits and a fly that is
homozygous recessive for both traits.
Collect a large number of F1 flies from the
cross. Determine the number of flies with
each of the possible phenotypes.
Calculate the total number of recombinant F1
phenotypes as a percentage of the total
number of F1 flies. This gives you the
recombination frequency and the number of
map units separating the two genes.
PpVv × ppvv
Possible non-recombinant phenotypes:
• purple eye, vestigial wing
• normal eye, normal wing
Possible recombinant phenotypes:
• purple eye, normal wing
• normal eye, vestigial wing
number of recombinant types
 100%
total number of offspring
Part A: Constructing a Chromosome Map
Procedure
1. Using the table above as a guide, determine the map distance between the linked genes for eye
colour and wing type in the following experiment. You perform the following cross: PpVv ×
ppvv. You count 1000 offspring in the F1 generation. You find that 406 of the flies have normal
eyes and normal wings, 398 have purple eyes and vestigial wings, 96 have normal eyes and
vestigial wings, and 100 have purple eyes and normal wings.
2. From previous research findings, you know that the distance between the gene for eye colour and
the gene for body colour is 12.2 map units, and the gene for body colour is 7.4 map units away
from the gene for wing type. All three genes are on the same chromosome. Draw a chromosome
map showing the relative distances between the three linked genes.
Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved.
This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher.
CHAPTER 17
HANDOUT
BLM 17.2.3
Thought Lab 17.1: Mapping
Chromosomes (cont’d)
Analysis
1. You conduct the same cross again, but this time you get an almost exact 1:1 ratio of flies with
normal eyes and normal wings to flies with purple eyes and vestigial wings. There are no
recombinant types. Provide two explanations that might account for these results.
2. Linkage data has been used to map genes on the chromosomes of Drosophila melanogaster. Do
you think such data could be used to map human chromosomes? Explain your reasoning.
Part B: Using Linked Gene Notation
In order to describe linked genes, fruit fly geneticists developed a different notation for genes. The
normal allele—the allele that is nearly always found in natural populations—is called the wild type
allele. The wild type allele is represented with a plus sign (+). The mutant allele for a gene is
represented with lower-case italic letters. Linked genes are represented with a horizontal line or a
slash.
For example, the gene for purple eyes (pr) and the gene for vestigial wings (vg) are on chromosome
number 2.
prvg
The homozygous recessive genotype is represented as
or pr vg / pr vg to indicate that these
prvg
genes are on the same chromosome.

The homozygous dominant genotype is represented as
or + + / + +.

Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved.
This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher.
CHAPTER 17
HANDOUT
BLM 17.2.3
Thought Lab 17.1: Mapping
Chromosomes (cont’d)
Procedure
1. Create a chromosome map of three linked genes based on the research presented below.
a) In fruit flies, the mutant gene d causes short legs and the mutant gene pr causes purple eyes.
A geneticist performs the following cross: pr d / + + × pr d / pr d. She counts 1000 offspring
and finds 391 wild type, 115 purple-eyed and normal-legged, 105 normal eyed and shortlegged, and 389 purple-eyed and short-legged. What is the map distance between the genes
for leg length and eye colour?
b) The same geneticist then performs the following cross: vg d / + + × vg d / vg d. She counts
1000 offspring and finds 350 wild type, 154 vestigial winged and normal-legged, 153
normal-winged and short-legged, and 343 vestigial-winged and short legged. What is the map
distance between the genes for leg length and wing type?
c) The recombination frequency for the gene for eye colour and the gene for wing type is 8.7%.
What is the map distance between these two genes? Draw a chromosome map showing the
relative distances between all three linked genes studied in Part B.
Analysis
1. Do you think the linked gene notation used by geneticists simplifies or complicates linked gene
analysis? Explain your reasoning.
Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved.
This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher.