Module 6 Lesson 4 - 4 (Lab)
Lesson 4 — Dihybrid Crosses
Lab: Mouse Genetics (Two Traits)
Introduction
Mendel wanted to know if the separation of alleles on one gene had any effect on the separation of alleles on another. To test this, he looked at plants that were pure breeding for two traits at once. He crossed plants that were true breeding
for two traits with plants that were true breeding for the opposite forms of the same traits.
In conducting and analyzing those crosses, Mendel was able to discover predictable patterns and ratios in the phenotypes of the F1 and F2 offspring.
This simulation explores mouse coat colour and eye colour as two separate genetic traits. You will conduct various breeding cycles until you understand how each trait separates in the F1 and F2 offspring. You will determine if the movement of alleles for one trait has any effect on the movement of alleles for the other trait.In this lab, the black fur colour (F) is dominant over white fur colour (f). The black eye colour (E) is dominant over red eye colour (e).
Problem (Purpose)
Manipulate the P1 breeding pair and observe the resulting offspring to determine which allele is dominant for each trait and to see if one trait has an effect on the other.
Materials
For this simulation, you require access to the Internet and a word processing program to record your results.
Procedure
Open the Gizmo Mouse Genetics
and the Exploration Guide. In this investigation, you will follow the instructions listed in the exploration guide for the parts
titled
- Patterns of Inheritance
- Independent Assortment
As you read and follow the instructions, be sure you can answer the questions listed.
Observations
Patterns of Inheritance
Follow all instructions under this heading. Stop after you have completed Step 4.
Assessment Questions
Complete the assessment questions in the simulation and check your answers.Background
Understanding the inheritance of two traits requires the learning of some basic probability.If two events are independent, the probability of both events occurring is equal to the probability of the first event multiplied by the probability of the second. For example, assume that the probability of wearing a white shirt and the probability of having spaghetti for dinner are independent. If there is a 20% chance of wearing a white shirt and a 30% chance of eating spaghetti, the probability of both occurring is 0.2 x 0.3 = 0.06, or 6%. (Of course, a person might want to avoid wearing a white shirt to a spaghetti meal. In that case, the probabilities are not independent.)
|
|
FE | Fe |
fE | fe |
| FE |
FFEE
Black Fur |
FFEe
Black Fur |
FfEE
Black Fur |
FfEe
Black Fur |
| Fe |
FFEe
Black Fur |
FFee
Black Fur, red eyes |
FfEe
Black Fur |
Ffee
Black Fur, red eyes |
| fE |
FfEE
Black Fur |
FfEe
Black Fur |
ffEE | ffEe |
| fe |
FfEe
Black Fur |
Ffee
Black Fur, red eyes |
ffEe
|
ffee
red eyes |
There are 12 genotypes with black fur and 4 genotypes with red eyes.
The same principle applies to genetics. Consider the case of breeding two FfEe mice. Based on Punnett squares, there is a 0.75 probability of an offspring with black fur (FF or Ff) and a 0.25 probability of an offspring with red eyes (ee). Thus, the probability of an offspring with black fur and red eyes is 0.75 x 0.25 = 0.1875, or 18.75% (3/16).
All possible offspring genotypes are shown on an expanded Punnett square above. Each parent donates one of four possible allele combinations to an offspring. This yields 16 possible offspring genotypes.
The probability of all possible genotypes are given below:
| FFEE: 1/16 = 0.625 | FfEE: 2/16 = 0.125 | ffEE: 1/16 = 0.625 |
| FFEe: 2/16 = 0.125 | FfEe: 4/16 = 0.25 | ffEe: 2/16 = 0.125 |
| FFee: 1/16 = 0.625 | Ffee: 2/16 = 0.125 | ffee: 1/16 = 0.625 |
Based on the genotype probabilities, calculate the corresponding phenotype probabilities:
- Black fur, black eyes (FFEE, FFEe, FfEE, FfEe): «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»16«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mn»2«/mn»«mn»16«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mn»2«/mn»«mn»16«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mn»4«/mn»«mn»16«/mn»«/mfrac»«mo»=«/mo»«mo»§#160;«/mo»«mfrac»«mn»9«/mn»«mn»16«/mn»«/mfrac»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»0«/mn»«mo».«/mo»«mn»5625«/mn»«/math»
- Black fur, red eyes (FFee, Ffee): «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»16«/mn»«/mfrac»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mfrac»«mn»2«/mn»«mn»16«/mn»«/mfrac»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mfrac»«mn»3«/mn»«mn»16«/mn»«/mfrac»«mo»§#160;«/mo»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»1875«/mn»«/math»
- White fur, black eyes (ffEE, ffEe): «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»16«/mn»«/mfrac»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mfrac»«mn»2«/mn»«mn»16«/mn»«/mfrac»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mfrac»«mn»3«/mn»«mn»16«/mn»«/mfrac»«mo»§#160;«/mo»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»1875«/mn»«/math»
- White fur, red eyes (ffee): «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»1«/mn»«mn»16«/mn»«/mfrac»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»0«/mn»«mo».«/mo»«mn»625«/mn»«/math»