Ch. 12� Bi - two� Centro - the center� Chroma - colored� Cyclo - a circle� Gamet - a wife or

husband� Gen - to create,

produce� Inter - in between

� Mal - bad, evil� Meio - less� Mere - a part� Meta - between� Mito - thread� Pro - before� Soma - body� Telos - an end

Plans for the week� Tuesday – linked genes� Wednesday – mutations to know and� Thursday – DNA structure &

replication� Friday – videos, discussion, diagrams

Challenging Genetics Practice Problems

11 problemsYou and a partner or partners will

work them together and report the answer and show the

solution on the screen

Ch. 12Linkage Groups & Chromosome Maps

When Mendel crossed his F1 generation: PpRr x PpRr, he got a 9:3:3:1 ratio. He would have not seen this pattern if the alleles had been located on the same chromosome and inherited together.

Objectives

1. Define and explain how we get parental and recombinant types

2. Create a chromosome map based on crossing over/recombinant types data

Chromosome Theory of Inheritance

1. Genes have specific locations on chromosomes

2. Genes on homologous chromosomes segregate away from each other during meiosis

3. Each gene pair segregates independently of other gene pairs

Thomas Hunt Morgan studied fruit flies and found that in some crosses, expected outcomes weren't happening. Further experiments confirmed that alleles located on the same chromosome are often inherited together. (i.e. they DON’T follow the law of independent assortment)

A common cross used to demonstrate linkage groups is the cross of a heterozygote wild type vestigial wings/ black body with a recessive mutant.

What is up with all those letters?

or just use the old fashioned way

A a B b x a a b b

Because these alleles are found on the same chromosome, a Punnett square would show them inherited together. There are two possible arrangements for the heterozygote (AaBb) in the above cross.

If the dominant alleles AB are on the same chromosome, it is called a CIS arrangement

If the dominant alleles are on different chromosomes (Ab) then it is called TRANS

Consider a cross with a parent that has the CIS arrangement to one that is a mutant for both traits (aabb)

a

b

a

b

A

B

a

b X

50% wild

50% mutants

Thomas Hunt did not observe a perfect 1:1 (or 50/50) ratio. Instead, his results looked like this….

Expected Observed

Wild Type 50 40

Mutant 50 40

Vestigial wings, Wild 0 10

Black body, Wild 0 10

Question: How would you explain these results?

Answer: The two offspring that did not look like either parent are called recombinants. They are a result of a CROSS-OVER that occurred during meiosisThe alleles switched position.

Parental and Recombinant TypesParental Types – offspring have the same phenotype as one of the two parents

Recombinant Types – offspring with a new combination of phenotypes

If recombinant phenotypes are seen 50% of the time, the genes are not linked (they are on separate chromosomes)

If some other pattern appears (less than 50%), the genes are linked

How often you see recombinants (the result of crossing over thus “unlinking” the genes) allows you to calculate the distance between any 2 genes on a chromosome

1. What is a parental type?

2. What is a recombinant type?

3. How do we get recombinant types?

4. How do we get parental types?

Practice Questions (assume no crossing over occurs)

1. A dumpy winged (dd) fruit fly with long aristae (AA) is crossed with a long winged (Dd) short aristae (aa). Show the cross and the phenotypic proportions.

2. A fruit fly with short legs (ll) and vestigial wings (ww) is crossed with one that is heterozygous for both traits. Assuming the dominant alleles are on separate chromosomes, show the cross and the expected phenotypic proportions.

3. In fruit flies, red eyes is a dominant allele located on the X chromosome. The recessive condition results in white eyes. The tan body trait is also X-linked and is dominant to yellow bodies. A female who is heterozygous both traits with the dominant alleles located on the same chromosome is crossed with a white eyed, yellow bodied male. Show the cross and the phenotypic proportions (Don't forget these traits are X-linked!)

4. Chromosome Map Problem

In pea plants, flower color and pollen shape are located on the same chromosome. A plant with purple flowers and long pollen (AaBb) is crossed with one that is recessive for both traits (aabb).

The results are as follows:

Results

Purple, long 47

Red, round 47

Purple, round 3

Red, Long 3

a) Are the chromosomes of the AaBb parent in the cis or trans position? Sketch the punnett square showing the expected offspring.

b) How far apart are the two alleles?

116

5. In minions, the phenotype for number of eyes is governed by two alleles – one eye = (E) and two eyes = (e). Two heterozygous minions mate to produce 465 offspring.Calculate how many of these offspring are expected to have two eyes. Round your response to the nearest whole number.

6.

In corn, purple kernels (R) are dominant to yellow kernels (r). Cobs from the offspring of a cross between a purple plant and yellow plant were used in a lab. A student counts 329 purple and 299 yellow kernels on one cob.Calculate the chi-squared value for the null hypothesis that the purple parent was heterozygous for purple kernels. Give your answer to the nearest tenth.

7.

A plant geneticist is investigating the inheritance of genes for bitter taste (S) and explosive rind (e) in watermelon. Explosive rind is recessive and causes watermelons to burst when cut. Non-bitter watermelons are a result of the recessive genotype (ss). The geneticist wishes to determine if the genes assort independently. She performs a testcross between a bitter/nonexplosive hybrid and a plant homozygous recessive for both traits. The following offspring are produced:

bitter/non-explosive – 88bitter/explosive – 68

non-bitter/non-explosive – 62non-bitter/explosive – 82

Calculate the chi-squared value for the null hypothesis that the two genes assort independently. Give your answer to the nearest tenth.Are these genes linked? How do you know?

8.

In DNA, A pairs with ___ and G pairs with ___So…if the genome of a fruit fly contains 2378 thymine nucleotides, how many adenine, guanine, and cytosine nucleotides would you expect if there are 4000 base pairs?

2738 adenine1622 cytosine1622 guanine

9.

Are these genes linked? How do you know?

10.

Is the student correct?

11.