1. A large ear of corn has a total of 433 grains, including 271 Purple & starchy, 73 Purple & sweet, 63 Yellow & starchy, and 26 Yellow & sweet.
Calculate the chi-squared value for the null hypothesis that the ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two genes for color and sweetness. Explain the chi-square test results.
So, your tentative hypothesis is: This ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1.
Test your hypothesis using chi square and probability values.
Total Kernals = 840
Expected values are as follows:
Purple and Smooth 9/16 of 433 = 243.6
Purple and Wrinkled 3/16 of 433 = 81.2
Yellow and Smooth 3/16 of 433 = 81.2
Yellow and Wrinkled 1/16 of 433 = 27.1
3.08 + .82 + 4.07 + .04 = 8.01
there are 4 phenotypic classes, so you have 3 degrees of freedom (4-1), and we want to be 95% accurate, so we look at the 0.05 column...
This is a poor fit since 8.01 > 7.81.
2. In a certain reptile, eyes can be either black or yellow. Two black eyed lizards are crossed, and the result is 72 black eyed lizards, and 28 yellow-eyed lizards.
Calculate the chi-squared value for the null hypothesis that the black eyed parents were heterozygous for eye color. Explain the chi-square test results.
Your Tentative Hypothesis: The black eyed parents were Bb x Bb.
Expected values:
If you do a punnet square of Bb x Bb, you get an expected percentage of 25% and 75% or 1/4 and 3/4
Since the total number of lizards is 100, the observed is already in the correct ratio
Expected ....................... Observed
Black Eyes = 75 ................ 72
Yellow Eyes = 25 ............. 28
Test your hypothesis using chi square analysis. In this set, because only two values (traits) are examined, the degrees of freedom (df) is 1.
.12 + .36 = .48 which is less than 3.84 (from the above table)
Good FIT
3. Problem: A sample of mice (all from the same parents) shows
· 58 black hair, black eyes
· 16 black hair, red eyes
· 19 white hair, black eyes
· 7 white hair, red eyes
a) What were the genotypes and phenotypes of the parents?
b) Use a chi square-test to support your hypothesis.
Looks like a 9/3/3/1 ratio, meaning the parents are HhEe x HhEe
Total number of offspring = 100
Expected values would be:
9/16 of 100 = 56.25
3/16 of 100 = 18.75
3/16 of 100 = 18.76
1/16 of 100 = 6.25
4. A wild-type fruit fly (heterozygous for gray body color and normal wings) was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution: wild type, 778; black-vestigial, 785; black-normal, 158; gray-vestigial, 162.
Calculate the recombination frequency (%) between the linked genes for body color and wing type.
First count the total number of offspring 778+785+158+162 = 1883
In all dihybrid test crosses (a cross between a known heterozygote for two wild type traits and a homozygous recessive individual for both traits) the expected ratio of phenotypes if the genes are on separate chromosomes must be:
wild type, 25%; black-vestigial, 25% black-normal, 25%; gray-vestigial, 25%. These results do not fit the experimental data above (778+785+158+162).
In fact the black-normal (158) and gray-vestigial (162) offspring represent recombinant individuals.
Calculation of recombination frequency:
778 - wild type
785 - black-vestigial
158 - black-normal
162 - gray-vestigial
778/1883 = 41.3%
785/1883 = 41.7%
158/1883 = 8.4%
162/1883 = 8.6%
83% are non-recombinant
17% are due to recombination
Recombination frequency = 17%
5. In another cross, a wild-type fruit fly (heterozygous for gray body color and red eyes) was mated with a black fruit fly with purple eyes. The offspring were as follows: wild-type, 721; black-purple, 751; gray-purple, 49; black-red, 45.
Calculate the recombination frequency (%) between the linked genes for body color and eye color.
Determine the recombination frequency:
Total flies = 1566
The percent recombination is therefore 6%
6. What fruit flies (genotypes and phenotypes) would you mate to determine the sequence of the body color, wing shape, and eye color genes on the chromosomes?
Example: cross a wild type fly heterozygous for body color and wing shape with a fly homozygous recessive for the same traits (black body/curly wings)
7. Determine the sequence of genes along a chromosome based on the following recombination frequencies: A-B, 8 map units; A-C, 20 map units; A-D, 25 map units; B-C, 12 map units; B-D, 33 map units.
C-B-A-D