Genetics Beyond Mendel

Lecture 3 exam 2

UIC BioS 101 Nyberg

Hypothesis Testing
Solving Genetics Problems

Multiple Alleles at a locus

Heterozygotes with a new Phenotype

Interactions among loci

Speaker Notes:

Many new properties of genes emerged as more species were studied. There can be many alleles at a locus. Heterozygotes may have a different phenotype than either homozygote. Alleles at one locus can control the expression of genotypes at another locus.

Lecture 3 exam 2

UIC BioS 101 Nyberg

Reading Assignment

You have already read Chapter 13. Now it is time to use your knowledge on genetics problems.
On p.292-293 try #11, 12, 13, 15, 18, 21 & 25.
There are some practice problems posted in the Exam 2 section of the website.

Speaker Notes:

You need to work with genetics problems to learn if you understand the ideas.

Lecture 3 exam 2

UIC BioS 101 Nyberg

Mendel’s Hypotheses

One allele is DOMINANT and the other is RECESSIVE in the heterozygote.
There are only two alleles at a locus.
Alleles segregate in gamete formation.
Loci assort independently at meiosis.
Genes are not changed in heterozygotes.

Speaker Notes:

Segregation is the only rule that has no exceptions.

Lecture 3 exam 2

UIC BioS 101 Nyberg

Phenotype of Heterozygote

The phenotype of a heterozygote is not always like one of the homozygotes.
- The heterozygote can manifest attributes of both alleles.
- The heterozygote can be a blend of the two homozygotes.
- The heterozygote can be completely different than either heterozygote, Hybrid Vigor.

Speaker Notes:

The distinction between codominance and incomplete dominance is not clear in many cases.

Lecture 3 exam 2

UIC BioS 101 Nyberg

Hybrid Vigor

Hybrids between lines may be more vigorous than their inbred parents.
Corn that is raised by farmers is a hybrid of a hybrid produced by seed companies. This hybrid corn has much higher yield.
Hybrids in animals, such as dogs, avoid breed specific problems, but also lack breed specific desired features.

Speaker Notes:

Corn is the clearest example where hybrids are preferred. For animals establishing breeds by inbreeding is the tradition.

Lecture 3 exam 2

UIC BioS 101 Nyberg

The ABO Blood Types

Alleles Genotypes Phenotypes

IA IA IA A

IB IA IB AB

i IA i A

IB IB B

IB i B

i i O

Speaker Notes:

Humans are polymorphic for these blood types. They are crucial for transfusions.

Three alleles mean 6 possible genotypes. In this case those 6 genotypes produce 4 phenotypes.

Lecture 3 exam 2

UIC BioS 101 Nyberg

One RULE in doing genetics problems is that a genotype has one and only one phenotype. For each possible genotype, given the alleles you hypothesize, you must assign a single phenotype.

Speaker Notes:

This rule is simple but it some students find it difficult to apply.

Lecture 3 exam 2

UIC BioS 101 Nyberg

An ABO genetics problem

If an individual with AB blood mates with an individual with O blood, what offspring blood types are possible and in what proportion are they expected?

AB individuals produce ½ IA and ½ IB gametes.

O individuals produce all (100%) i gametes.

IB i

IA i

Speaker Notes:

The box on the left shows the gametes and the genotypes of the resulting zygote. The box on the right translates those genotypes into phenotypes.

Lecture 3 exam 2

UIC BioS 101 Nyberg

More ABO genetics

A student has type A blood. Both her parents have type A blood. What observation would tell you that BOTH of her parents were heterozygous, IAi ?
ANSWER If any of the student’s siblings had type O blood.

Speaker Notes:

If either parent were homozygous, ALL the offspring would have the A phenotype.

If three children exist an all have type A blood, it is still quite possible that both parents are heterozygous.

Lecture 3 exam 2

UIC BioS 101 Nyberg

A Drosophila Cross

What do you infer from the statement that lines X and Y are true-breeding lines?
One line had WHITE eyes, the other RED. A male from WHITE line was crossed to a female from the RED line. The F1 had RED eyes. What genetic model would you propose that explains the available facts?

Speaker Notes:

The statement that a line is true-breeding implies that the line is homozygous. The cross between two true-breeding lines results in all heterozygous offspring (like Mendel’s F1).

Lecture 3 exam 2

UIC BioS 101 Nyberg

A Drosophila Cross

The available information suggests a situation similar to that Mendel observed for seed color.
Hypothesis #1:
- One locus, two alleles (W = Red, w = White).
- Red allele, W, dominant to white allele, w.
PREDICTION of F2 results.

Speaker Notes:

You have to make up symbols for genes when you know only phenotypes. The classic way is to develop a symbol from the recessive trait.

Lecture 3 exam 2

UIC BioS 101 Nyberg

A Drosophila Cross

Hypothesis #1 (one locus two alleles) predicts that there will be two phenotypes in the F2;
RED = 0.75
WHITE = 0.25

The observed results were RED =390. WHITE = 110.

Speaker Notes:

The results are compatible with the ¾ RED and ¼ WHITE hypothesis for chi-squared = 2.62 which is less than 3.84. BUT a careful observer noticed that ALL 110 WHITE eyed flies were males.

Lecture 3 exam 2

UIC BioS 101 Nyberg

A Drosophila Cross

Hypothesis #2.
Some student noticed that there is a locus, white, on the X chromosome that results in a WHITE eye when homozygous.
Both male and female F1 flies have red eyes. This is possible only if the original cross was WHITE male by RED female.

Speaker Notes:

YES when the student looked at the records the original cross involved males from the WHITE line and females from the true-breeding RED line.

Our original cross under hypothesis 2

Lecture 3 exam 2

UIC BioS 101 Nyberg

Speaker Notes:

Figure 13.11a p277

Lecture 3 exam 2

UIC BioS 101 Nyberg

A Drosophila Cross

Hypothesis 2 continued.
The RED F1 male would produce two kinds of gametes, Y and W (or w+ ).
The RED F1 females would produces two kinds of gametes, W (or w+ ) and w, in equal proportions.

Speaker Notes:

Males have the XY chromosome arrangement. When they undergo meiosis they produce half gametes with a Y chromosome and half with the X chromosome. The Y chromosome carries no information except it determines maleness. The W and w alleles are only on the X chromosome.

Segregation of sex chromosomes in formation of sperm

Lecture 3 exam 2

UIC BioS 101 Nyberg

Speaker Notes:

Figure 13.10 p277

Lecture 3 exam 2

UIC BioS 101 Nyberg

A Drosophila Cross

RED

male

WHITE

male

RED

female

RED

female

Hypothesis #2. WHITE eye color due to white gene on X chromosome. Predicted F2:

Speaker Notes:

This hypothesis also predicts 3/4th RED and ¼th WHITE, but all the WHITE flies are male.

Lecture 3 exam 2

UIC BioS 101 Nyberg

Comparisons of Hypotheses

In both hypotheses the predicted F2 flies are expected to be 75% RED and 25% WHITE. In hypothesis #1 the RED and WHITE eye colors should appear likely among males and females. In hypothesis #2 100% of the females will have RED eyes, but 50% of the males will have RED eyes, and 50% will have WHITE eyes.

Speaker Notes:

The observed results indicate that hypothesis 2, sex linkage if the white allele is the correct hypothesis.

Lecture 3 exam 2

UIC BioS 101 Nyberg

EPISTASIS
GENE INTERACTIONS

There are many loci that can effect eye color in Drosophila.
As is the case with the combs of chickens described in the book (p. 284), interactions among genotypes at different loci can be complex.
We will study a case of coat color in rodents.

Speaker Notes:

In epistasis the genotype at one locus is affected by the genotype at another locus.

Lecture 3 exam 2

UIC BioS 101 Nyberg

In mice a cross between a true-breeding BROWN and a true-breeding ALBINO yielded F1 individuals that were BROWN.

The simple hypothesis is that there is a recessive albino allele, a, and a dominant brown allele A. This hypothesis predicts that the F2 will have 75% BROWN and 25% ALBINO.

The observed results included a new phenotype, BLACK. A new phenotype is not possible with a one locus model.

Thus we are forced to consider the next level of model, a two locus model.

What were the observed RESULTS? N = 100

BROWN 58

BLACK 18

ALBINO 24

A complex mouse cross

Speaker Notes:

The appearance of a phenotype that was not predicted instantly disproves a tentative genetic hypothesis.

Lecture 3 exam 2

UIC BioS 101 Nyberg

Here is the solution that was worked out:

Parents BROWN BBAA ALBINO bbaa

Gametes BA ba

F1 BbAa BROWN

Gametes

BA Ba bA ba

BA BR BR BR BR

Ba BR A BR A

bA BR BR BL BL

ba BR A BL A

F2 results

Speaker Notes:

The BL symbol means the BLACK phenotype. In the F2 3/16th on the progeny are BLACK, ¼ will have the ALBINO phenotype (symbolized by A in the slide).

Lecture 3 exam 2

UIC BioS 101 Nyberg

In this model, there is a locus albino with two alleles, A and a. The homozygote aa always has the ALBINO phenotype. Individuals with either the Aa or the AA genotype don’t have a specific coat color without reference to at least one other locus. The AA and Aa genotypes just allow the coat color alleles at other loci to be manifest. In this case the other locus has two alleles, B and b. The bb genotype is BLACK and the BB and Bb are BROWN given that there is at least one A allele at the A locus.

Two loci of coat color

Speaker Notes:

Genotype and phenotype relations can get quite complex.

Lecture 3 exam 2

UIC BioS 101 Nyberg

Take Home Problem

In the mouse coat color problem just discussed, what proportions of BLACK, BROWN and ALBINO coats are expected in the F2 generation?
What proportion of the F2 individuals are expected to be heterozygous at the albino locus?

Speaker Notes:

Count up the phenotypes from the expected genotypes in the F2. Total the proportions of each of the three genotypes.

Lecture 3 exam 2

UIC BioS 101 Nyberg

Vocabulary

Epistasis
Hybrid vigor
Rules of genetics
True-breeding

Quantitative prediction
New phenotype
Interactions among loci
Sex linkage

Speaker Notes: