Imagine that you attempted to recreate Mendel's work with garden peas. You began by crossing true breeding violet-flowered, tall plants with white-flowered dwarf plants.

After self-crossing the F1 generation, you obtain the following phenotypes in the F2generation:

80 tall, violet flowers
36 tall, white flowers
39 dwarf, violet flowers
5 dwarf, white flowers

Part A

How many tall plants with violet flowers were expected?

Express your answer as a whole number.

Part B

How many tall plants with white flowers were expected?

Express your answer as a whole number.

Part C

How many dwarf plants with violet flowers were expected?

Express your answer as a whole number.

Part D

How many dwarf plants with white flowers were expected?

Express your answer as a whole number.

Part E

Use a chi-square analysis to test the hypothesis that the F2 data for stem length (tall:dwarf) is consistent with Mendel's law of segregation.

Calculate the chi square value.

Express your answer to two decimal points (example: 100.00).

Part F

Can the deviation of observed values from expected values be attributed to random chance?

yes or NO

Use the probability chart to answer this question.

Use the probability chart to answer this question.

yes
no
Part G

Use a chi-square analysis to test the hypothesis that the F2 data for stem length (tall:dwarf) and flower color (violet:white) is consistent with Mendel's law of independent assortment.

Calculate the chi square value.

Express your answer to two decimal points (example: 100.00).

Part H

Can the deviation of observed values from expected values be attributed to random chance?

Use the probability chart to answer this question.

Use the probability chart to answer this question.

no
yes

A. 90

B. 30

C. 30

D. 10

E. 0.53

F. Yes

G. 7.51

H. Yes

Explanation:

There are a total of 160 plants (80   +   36   +   39   +   5 = 160).

In Mendel's experiment, an F1 generation of a dihybrid cross between two true breeding plants gives 100% heterozygous plants. We self-cross them to create the F2, in which we expect a ratio of 9:3:3:1. The dominant parental phenotype would be 9, the two recombinant phenotypes would both be 3, and the recessive parental phenotype would be 1.

Therefore, we calculate the total number of plants divided by the total number of potential outcomes from the cross (16), multiplied by 9, which is the expected number of tall violet plants

160 /  16 = 10

10 x 9 = 90

We carry out this calculation as above, within Mendel's law of 9:3:3:1, tall plants with white flowers represent the recombinant phenotype, meaning we divide the total number of flowers by the total number of potential outcomes from the cross (160/16 = 10). We then multiply this by the expected ratio, which is 3

10 x 3 = 30

This phenotype is also a recombinant phenotype, meaning the expected ratios are as in part B :  dwarf plants with violet flowers represent the recombinant phenotype, meaning we divide the total number of flowers by the total number of potential outcomes from the cross (160/16 = 10). We then multiply this by the expected ratio, which is 3

10 x 3 = 30

Dwarf plants with white flowers are the recessive parental phenotype, within the ratio of 9:3:3:1, they represent 1. Again, we divide the total number of flowers by the total number of potential outcomes from the cross (160/16 = 10). We then multiply this by the expected ratio, which is 1

10 x 1 = 10

Mendel's law of segregation states that alleles segregate randomly into gametes. To assess whether the law of segregation is being adhered to here, in terms of stem length, we perform a chi squared test, which has the formula

Sum of = for all classes

Where O is the observed value and E is the expected value. The easiest way to do this is to construct a simple table:

The observed number of tall flowers is 80 + 36 = 116, the expected number is 90 + 30 = 120. The observed number of dwarf flowers is 39 + 5 = 44. The expected number of dwarf flowers is 30 + 10 = 40

                                               O             E          (O-E)²          (O-E)²/ E

tall flowers              .              116           120          16                0.13

dwarf flowers                        44           40            16                0.4

TOTAL                                                                                     = 0.53

To see whether the deviation of the observed values from the expected values can be attributed to random chance, we first have to know the degrees of freedom.

The degrees of freedom are the number of tested outcomes - 1. In this case there are 2 phenotypes, so 1 degree of freedom.

We then consult a table showing the critical values of the chi squared distribution, to see the maximum chi squared value for our degrees of freedom, within our threshold for statistical significance. A table is attached. The table tells you the probability that the chi squared value was exceeded by random chance.

Our value of 0.53 corresponds to a p value of more than 0.5. This means that, if the law of segregation is true, then deviations from expected values this large are expected approximately 50 percent of the time. our level for significance is 0,05 (5%), so the null hypothesis cannot be rejected and the law of segregation applies.

Mendel's law of independent assortment states that any combination of alleles from the parents can be inherited in the offspring. I.e. the offspring have just as much chance of inheriting both tall white alleles as both dwarf violet alleles.

Sum of = for all classes

                                               O             E          (O-E)²          (O-E)²/ E

tall, violet flowers                 80           90          100                 1.11

tall, white flowers                 36           30           36                  1.2

dwarf, violet flowers           39           30            81                   2.7

dwarf, white flowers           5             10           25                   2.5

TOTAL                                                                                     = 7.51

The degrees of freedom are the number of tested outcomes - 1. In this case there are 4 phenotypes, so 3 degrees of freedom.

Our threshold for significance (p value) is 0.05 unless told otherwise. Our chi squared value corresponds to a p value between 0.1 and 0.05, meaning that deviations from expected values this large are expected approximately 5-10% of the time. This is within our significant threshold, of 0.05 (5%) so our deviations can be attributed to random chance.

an individual’s sex (i.e., whether they are a male or female) is determined by the sex chromosomes. most people have two sex chromosomes, one that is inherited from their mother and one that is inherited from their father. typically, females have two x chromosomes (xx) and males have one x chromosome and one y chromosome (xy). conditions caused by changes (“mutations”) in genes located on the x chromosome are considered x-linked.

x-linked recessive inheritance

most x-linked conditions are recessive. this means that in a person with two x chromosomes (most females), both copies of a gene (i.e., one on each x chromosome) must have a change or mutation whereas in a person with one x chromosome (most males), only one copy of a gene must have a mutation. a female with a mutation in one copy of a gene on the x chromosome is said to be a “carrier” for an x-linked condition. a male with a mutation in a gene on the x chromosome is typically affected with the condition. because females have two copies of the x chromosome and males have only one x chromosome, x-linked recessive diseases are more common among males than females. however, x-linked recessive diseases can occur in both males and females.

for x-linked recessive disorders, an unaffected carrier mother who has a mutation in a gene on the x chromosome can transmit either the x chromosome with this mutation or a “normal” x chromosome to her children. if the father is unaffected, none of her daughters will be affected and all of her daughters will be unaffected—since they will inherit at least one normal x chromosome from their father. however, each daughter will have a 50% chance of being an unaffected carrier like her mother and a 50% chance of both x chromosomes being normal.