Mercer Community College

Mercer County Community College

Division of Science and Health Professions

BIO 208

Genetics

Applied Human Genetics

Single Gene (Monohybrid) Crosses

Probability

Chi Square Analysis

Applied Human Genetics/Single Gene Traits

The classical study of single gene traits is by pedigree analysis. Over 7000 human single gene traits have been recognized, many also assigned to the particular chromosome on which they reside. The Human Genome Project will uncover many more genes and further study in proteomics will assign function to these genes. Chromosome 22 contains over 500 genes and the larger chromosomes contain thousands of genes. For more information on the mechanism of inheritance of single gene traits, allelic variations, and chromosome mapping, consult the Online Mendelian in Man web site at www.OMIM.gov.

The genotype of a homozygous recessive individual is usually known by the phenotype exhibited. For example, individuals with blue eyes are bb. However, the dominant phenotype could arise from two genotypes. A person with the dominant brown eye phenotype could have the BB or Bb genotype. Therefore the genotype is represented as B-. The dash indicates that while a second allele is present, the nature of that allele is unknown

BB = homozygous dominant

Bb = heterozygous B- indicates dominant phenotype

bb = homozygous recessive

Single gene traits

Interlocking fingers (I)

Interlock your fingers and observe which thumb is on top (right or left). The tendency to place the left thumb on top is due to a dominant allele (I). Those who place the left thumb on top have the genotype I- (either II or I-). The right thumb on top is determined by the ii genotype

Ear lobes (E)

Examine your partner’s earlobes. The dominant allele (E) results in the phenotype of free earlobes. The recessive allele (e) is for attached earlobes. Earlobes are attached if the bottom lobe is attached directly to the head. Earlobes are free if the lobe hangs free.

Widow’s peak (W)

Widow’s peak occurs when the hairline forms a distinct point in the center of the forehead. Lift up the hair of your forehead to score this trait. People without widow's peak have a smooth hairline with no dip. Men starting to go bald (or already bald) may be unable to score this trait. Widow's peak is controlled by a dominant allele (W)

Tongue curling (T)

A dominant allele (T) results in the ability to curl the tongue in a U-shape

Hitch hiker’s thumb (H)

A person homozygous recessive for this trait (hh) can bend the last (distal) thumb joint back to about a 90 degree angle. Those with the H allele cannot

Pigmented iris (B)

A person with the dominant allele has brown or brownish eyes. The recessive allele encodes blue eyes

Polydactyly (PD)

A person with polydactyly has >5 fingers on 1 or more hands and feet. It is dominant (PD)

Mid-digital finger hair

Examine middle segment of your fingers. If hair is present, even one, you have mid-digital hair.

PTC tasting (P)

If you can taste the chemical PTC, you have a dominant allele for this trait (P). Place the strip of PTC paper on your tongue for a few seconds. If you cannot taste anything, you do not possess the dominant allele

Fill in the following table

Gene / Describe phenotype / Genotype
Earlobes (E)
Widows peak (W)
Tongue roll (R)
Hitchhikers thumb (H)
Pigmented iris (P)
Polydactyly (PD)
PTC taster (T)
Interlocking fingers (I)

Class phenotypic frequencies.

Gene / f (E-) dominant phenotype / f(ee) recessive phenotype
Tongue roll (R)
Pigmented iris (P)
Polydactyly (PD)
PTC taster (T)
Interlocking fingers (I)

Maize Genetics: Single gene crosses

Zea mays is a model organism in genetics. A large number of progeny are produced from each cross as each kernel is a single individual. All kernels on one ear represent the progeny of a single cross. The effects of many genes can be observed in the kernels, eliminating the need to grow plants. Many genes determine the phenotypes of the 3 tissues that control the color of a corn kernel.

Endosperm (kernel) traits

Color gene Kernel shape gene

P allele = purple (color) R allele= smooth

p allele = yellow (no color) r allele= wrinkled

STATION 1. Obtain an ear of corn from the instructor

1. What are the two color phenotypes observed?

2. Count 50 kernels. Compute the ratio of phenotypes:

3. Which phenotype is caused by a dominant allele?

4. Show the cross of the genotypes and phenotypes of the parental plants that led to this cob.

Parental genotypes à ______X ______

Parental phenotypes à ______

STATION 2: Obtain an ear of corn from the instructor.

1. What are the two kernel shape phenotypes observed?

2. Count 50 kernels. Compute the ratio of phenotypes:

3. Which phenotype is caused by a dominant allele?

4. Show the cross of the genotypes and phenotypes of the parental plants.

Parental genotypes à ______X ______

Parental phenotypesà ______

STATION 3: The ear of corn represents the progeny of a cross between two purple, smooth parents.

Genotype: PpRr X PpRr

Phenotype: Purple, smooth Purple, smooth

1. What are the 4 allelic combinations of gametes (eggs and sperm) each parent can produce? Remember, a gamete can only and must contain one allele of each gene.

2. Construct a Punnett square showing the expected genotypes and phenotypes of the offspring

3. What are the 4 phenotypes (with respect to both kernel shape and color) observed in the offspring of this cross?

4. Count 100 kernels. Compute the ratio of phenotypes (divide each number by the smallest number)

5. Does this observed ratio approach the expected ratio?

Chi Square

The purpose of the Chi Square (X2) test is to determine if data obtained experimentally (observed data) fit into a model of expected data. If data are shown to fit into the model, then any variations from the model are due to chance. If the observed data do not approximate the model, then variations are said to be significant, or not due to chance. If deviations are small they are more easily attributed to chance than are large ones.

X2 = S ( O – E ) 2

E O observed data

E expected data

Chi Square values have been published by statisticians in a table (below). Across the top of the table are probability values (P). A P value >0.05 indicates that in over 5% of trials, similar data would be obtained. A p >0.05 leads the researcher to accept the null hypothesis, that the differences between observed and expected data are due to chance. In other words, the observed data fits the expected data.

A P value <0.05 corresponds to a 1 chance in 20 (5%) that deviations are due to significant variations. A p 0.05 indicates that observed variations cannot be explained by the expected model. In other words differences in the observed data are not due to chance.

The null hypothesis states that any variations between observed and expected data are only due to chance. Rejecting the null hypothesis indicates that observed and expected data are significantly different.

To determine which row of Chi Square values to use, first determine the degrees of freedom. The degrees of freedom are one less than the number of phenotypic classes. For example if white, yellow, purple, and red corn kernels are observed, the degrees of freedom is 3.

I. Application of data: Maize

A. Enter the data obtained from ear 1

Phenotype / Observed O / Expected E / Deviations
(O-E)2 / (O-E)2
E
TOTAL

X2 = ______

1. How many degrees of freedom (df) are there?

2. Locate the two Chi Square values on either side of the calculated X2. What is the lowest probability value assigned to these X2s?

p =

3. Do you accept the null hypothesis (variations between observed and expected data are only due to chance) or do you reject the null hypothesis (variations between observed and expected data are significantly different? Explain

4. In general, does a low Chi Square value support the null hypothesis?

B. Enter the data obtained from ear 3 (2 gene cross)

Phenotype / Observed
O / Expected
E / Deviations
(O-E)2 / (O-E)2
E
TOTAL

X2 = ______

1. How many degrees of freedom (df) are there?

2. What is the lowest probability value assigned to these X2s?

p =

II. Application of data: Beans

Obtain a handful of beans from a jar in which two different colors of beans have been mixed in equal amounts. Count each color and enter in the table below

Phenotype / Observed O / Expected E / Deviations
(O-E)2 / (O-E)2
E
TOTAL

X2 = ______

1. How many degrees of freedom (df) are there?

2. Locate the two Chi Square values on either side of the calculated X2. What is the lowest probability value assigned to these X2s?

p =

Probability

The coin toss is a convenient tool to demonstrate principles of probability. Probability in coin tossing involves chance. There is a ½ chance that a head will face up and a ½ chance that tails will occur. No matter how many times a coin is tossed, the probability of heads or tails remains the same.

If two coins are tossed at the same time, there are 4 combinations that can occur:

2 heads HH

1 head, one tail HT, TH

2 tails TT

The chance of obtaining two heads is determined by employing the product rule: the probability of two events occurring simultaneously is the product of their individual probabilities.

1. What is the probability of obtaining two heads? ______

The chance of obtaining a two heads OR a tail and a head is determined by using the sum rule: The chance of either of two events occurring is the sum of their individual probabilities. First use the product rule to determine the chance of HT and TH, then the sum rule to determine the chance of HT OR TH.

2. The probability of obtaining HH or TH is ______

3. Data collection

Obtain 3 coins.

a. What are the possible combinations of each class? Enter this information in the table

b. What is the probability of each class? Enter this information in the table

c. Toss the 3 coins 12 times. Combine your data with that of another group so that you have a total of 24 throws. Enter in the observed numbers column

Classes Combinations Probability______Observed numbers

3 heads ______

2 heads, 1 tail HHT, HTH, THH 1/8+ 1/8+ 1/8 = 3/8 ______

1 head, 2 tails ______

3 tails ______

Totals ______8/8 _____ 24_____

Binomial expansion

Expectations for combinations of outcomes in groups can be obtained mathematically by using the following binomial formula:

P = ______p x q n-x

x! (n-x)!

Note: 0! = 1

1. 5 babies are born. What is the probability that 2 are boys and 3 are girls?

n! = 5! = 5 X 4 X 3 X 2 X 1 n = number of events

x! = 2! = 2 X 1 x = number of event 1

(n-x)! = 3 X 2 X 1 n-x = number of event 2

p = probability of a boy = ½

q = probability of a girl = ½

Determine P (2 boys, 3 girls)

2. Albinism is a recessive trait. Those with the albino phenotype are aa. Heterozygotes are Aa, they are not albino, but are carriers for the trait. Two black mice, heterozygous for albinism were mated.

Aa X Aa

1. What is the probability of obtaining a white mouse?

2. A black mouse?

3. In a litter of 6 mice, determine P (3 white, 3 black)?

4. In a litter of 5 mice, determine P (all black)?