Define the following terms:

Genotype

Phenotype

Principal of segregation

Principal of independent assortment

Punnett Square

P1 generation

F1 generation

F2 generation

Alleles

Homozygous recessive

Homozygous dominant

Monohybrid cross

When do you get a ratio of 1:2:1?

Dihybrid cross

When do you get a ratio of 9:3:3:1?

Test cross: used to determine whether an individual showing the dominant trait is homozygous dominant or homozygous recessive

The individual that shows a dominant phenotype but is either homozygous dominant or heterozygous is crossed to a homozygous recessive individual

Probabilities:

The probability of two events BOTH occurring is the product of their individual probabilities = A and B occurring = (probability of A)(probability of B)

Can be used to calculate phenotypes of dihybrid and trihybrid crosses.

Yellow is dominant to green

Round is dominant to wrinkled

Cross YyRr X YyRr

Get 9:3:3:1

Look at each gene individually

Yellow: 1:2:1 or 3:1 = 1YY: 2Yy:1yy

Round: 1:2:1 or 3:1 = 1RR:2Rr:1rr

Problem

Cross YyRr X yyrr

What are the probabilities for each phenotype?

½ = Yy, ½ = yy

½ = Rr, ½= rr

½ yellow X ½ round = ¼ yellow and round

½ yellow X ½ wrinkled = ¼ yellow and wrinkled

½ green X ½ round = ¼ green and round

½ green X ½ wrinkled = ¼ green and wrinkled

Trihybrid cross

P1 = AABBCC X aabbcc

Gametes = ABC and abc

F1 = AaBbCc

Gametes = ABC, ABc, AbC, Abc, aBC, aBc, abC, abc

Cross F1 = AaBbCc X AaBbCc

Problem:

Determine genotype and phenotype for this cross

AaBBCc X aaBBCc

Cross As = ½ Aa, ½ aa

Cross Bs = all BB

Cross Cs = ¼ CC, ½ Cc, ¼ cc

Genotypes:

AaBBCC = (1/2)(1)(1/4) = 1/8; phenotype = A, B and C

AaBBCc = (1/2)(1)(1/2) = 1/4; phenotype = A, B and C

AaBBcc  = (1/2)(1)(1/4) = 1/8; phenotype = A, B and c

aaBBCC = (1/2)(1)(1/4) = 1/8; phenotype = a, B and C

aaBBCc = (1/2)(1)(1/2) = 1/4; phenotype = a, B and C

aaBBcc = (1/2)(1)(1/4) = 1/8; phenotype = a, B and c

total = 4/8 + 2/4 = ½ + ½ = 1

Problem:

AaBbCc X AaBBCC;  What are the probabilities for the different genotypes and phenotypes?

If n = # of heterozygote gene pairs

2ⁿ = the number of different gametes formed from each parent

3ⁿ = the number of different genotypes

2ⁿ = the number of different phenotypes

The probability of A or B occurring is the probability of A + probability of B

Conditional probabilities:  divide the probabilities

            Example:  what is the probability that a plant that is tall will also be heterozygous

            Probability of heterozygous/probability of being tall = 2/4/3/4 = 2/3

Tt X Tt = 1TT: 2Tt:1tt

Binomial theorem:  used to determine the probability that you will have X children who are male or female or who have a certain disease.

Probability = n!/s!t!(a^sb^t)

n = total # of events

s = the number of times a occurs

t = the number of times b occurs

a = probability of a

b = probability of b

What is the probability that of 4 children, 2 will be male and 2 will be female?

Probability = n!/s!t!(a^sb^t)

                        = (4!/2!2!)(1/2)²(1/2)²

                                        4X3X2X1  X  (1/2)²(1/2)²

                        2X1X2X1

                        = 6 X (1/2)⁴

                                        =  6 X ½ X ½ X ½ X ½ = 6/16 = 3/8

What is the probability that of 7 children, 5 will be male and 2 will be female?

Probability = n!/s!t!(a^sb^t)

                        = (7!/5!2!)(1/2)⁵(1/2)²

                                        = 7X6X5X4X3X2X1 X !)(1/2)⁵(1/2)²

                           5X4X3X2X1X2X1

                        = (7 X 6)/2 = 21 X ½ X/ ½ X ½ X ½ X/ ½ X ½ X ½

                        = 21/128

Problem

In a family of 5 children, what is the probability that 3 are male and 2 are female?

In a family of 4 children where both parents are heterozygous for cystic fibrosis (an autosomal recessive disease), what is the probability that 2 of the children are normal (have 1 or 2 dominant genes) and 2 have cystic fibrosis.

Statistics

If you have a large sample size, deviations from the expected numbers or ratios are less, there is less impact for chance to affect the results.

Null hypothesis:  there is no difference between the measured values (ratios) and the predicted values

If null hypothesis is rejected, the observed differences are not due to chance alone.  There is a difference between the different groups.

If you fail to reject the null hypothesis, then the deviations from expected values are due to chance alone.  There is no difference between your groups.

In statistics, p<0.05 means there is a statistical difference.  The groups are different and it is not due to chance.

Chi square test measures how frequently you observe the amount of deviation due to chance alone.

c² = sum(o-e)²/e

o = observed value

e = expected value

o-e = deviation

c² = d²/e

Degrees of freedom = n-1 (n = number of different categories or groups)

Monohybrid cross: 3:1 ratio = n = 2, df = 1

Dihybrid cross: 9:3:3:1 ratio = n = 4, df = 3

 With this information, you can calculate p and determine if your groups are statistically difference or if the difference is due to chance.

In both cases, p>0.05.  These groups are not different from the expected values.

The observed deviations from expected values can be attributed to chance.  We cannot reject the null hypothesis.

Problem

In a genetics laboratory, a student crossed flies with normal long wings to flies with mutant dumpy wings.  After the cross, there were 792 long-winged flies and 208 dumpy winged flies.

a)     do you think that dumpy wings was inherited as a dominant or recessive trait?

b)     perform a c² test to support or reject your hypothesis

p is between 0.01 and 0.001 based on table; the graph shows p = 0.001

The data shows that we can reject the null hypothesis.

The difference between the observed and expected is not due to chance alone.

Some of the flies may be less viable and do not support Mendel’s 3;1 ratio.

Problem

In a dihybrid cross, there were 315 round yellow peas, 108 round green peas, 101 wrinkled yellow peas and 32 wrinkled green peas.

Use a c² test to determine if this data fits the 9:3:3:1 ratio and whether or not you can reject or accept the null hypothesis.

Pedigree analysis

Pedigrees Reveal Patterns of Inheritance in Humans

Problem:  Why does this pedigree show autosomal dominant inheritance?  What are the genotypes of each person?