BIOL207 Lecture Notes - Lecture 18: Epistasis, Dihybrid Cross, Wild Type

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turquoisewalrus441

5 May 2018

School

University of Alberta

Department

Biology (Biological Sciences)

Course

BIOL207

Professor

Michael Harrington

For unlimited access to Class Notes, a Class+ subscription is required.

Mapping Human Genes.

1. Single Nucleotide polymorphisms (SNP) are used as genetic markers.

2. GEnome-Wide association studies (GWAS) where results are displayed with a

manhattan plot.

a. Single peak = 1 gene is involved → tells the relative location of gene on

chromosome.

b. Multiple Peaks = multiple genes are involved

c. Works with model and non-model organisms.

Maps used by geneticists

1. Cytogenetic → based on chromosome bands (3B6; not the same as humans!)

2. Genetic → based on meiotic recombination (1-1.5 = chromosome 1 1.5 mu from an end)

3. Sequence → based on DNA sequencing.

Incomplete Penetrance

❏Percent of individuals with a mutation that have the mutant phenotype.

Gene Interactions

1. If Dihybrid cross does not lead to an F2 of 9:3:3:1 there is gene interaction!

a. 9:3:3:1 means 2 genes, autosomal and on different chromosomes with no

interaction.

2. Complementary Genes → both genes needed to produce a wild type phenotype.

a. Modified F3 ratio of the 9:3:3:1 ratio means two genes, autosomal on different

chromosomes and there is gene interactions.

b. A and B genes in same circle and make one colour but if one absent make

intermediate.

3. Duplicate Genes → either of the two genes are needed to produce a wild type

phenotype.

a. Modified 9:3:3:1 ratio (example is 15/16 and 1/16).

b. A and B in same circle and one required for a different colour.

4. Discontinuous variation → character is found in two or more distinct forms in a

population. → Peas.

5. Continuous variation → when character is found in a range → humans

6. Epistasis → when one gene conceals another gene (note: Dominance talks about

alleles).

a. Recessive epistasis → When recessive phenotype of one gene conceals the

phenotype of a second gene.

i. Pigment and colour. (9:4:3 ratio)

ii. Linear pathway

b. Dominant epistasis → dominant phenotype of one gene conceals the phenotype

of another gene.

i. Onion bulbs (12:3:1 ratio)

ii. Branched pathways.

Determination of Genes

1. How many Genes?

2. Is it autosomal? → Sex linked? X or Z?

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Related Questions

Which of these statements is incorrect?

Syntenic genes are located on the same chromosome.

Independent assortment results in recombinant chromosomes.

You can reliably predict the relative genetic distance fromgenesâ physical distance on a chromosome.

Linked genes are always syntenic.

What is the relative genetic distance between two linked genesif the recombination frequency is 0.49?

0.49 cM

4.9 cM

49 cM

490 cM

What statement best explains the distortion in Mendelian ratiosobserved by Bateson & Punnett in 1905? (Reminder: they found anoverrepresentation of F2 offspring showing both dominant orrecessive phenotypes, and an underrepresentation of offspringdisplaying one dominant and one recessive phenotype)

Human error: they should have been more careful about theirexperimental setup.

Gene linkage: Genes for flower color and pollen shape arephysically close on the same chromosome, leading to a breakdown inthe independent assortment of the alleles for these traits.

Chromosome crossover: Homologous recombination of twochromatids during meiosis caused the alleles to shuffle, resultingin a breakdown of the independent assortment of the alleles forthose genes.

Random variation: No two situations are alike. In finitepopulations, you are going to get some variation across a mean.

When determining the relative genetic distance between twogenes, why is dihybrid back-cross preferable over traditionaldihybrid cross?

9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1ratio.

Genotypes of the offspring can be determined based on theirphenotype.

If the genes are independently assorted, the dihybrid back-crosswould result in only 2 genotypes in the F1 generation.

B and C

Why do we map genes?

To understand how genes interact with each other

Comparative genomics analysis

To determine the genotype of an organism

All of the above

coralox210

Please show detailed work. Thanks in advance.

16. Syntenic genes can assort independently when

A) they are very close together on a chromosome.

B) they are located on different chromosomes.

C) crossing over occurs rarely between the genes.

D) they are far apart on a chromosome and crossing over occurs frequently between the genes.

E) they are far apart on a chromosome and crossing over occurs very rarely between the genes.

17. The alleles of linked genes tend to

A) segregate together more often than expected by random assortment

B) assort independently.

C) be mutated more often than unlinked genes.

D) experience a higher rate of crossing over.

E) assort independently and show a higher rate of crossing over.

18. If you know that the frequency of recombination between genes X and Y is 34% and between X and Z is 25%, can you predict the order of the three genes?

A) Yes; the order is X-Z-Y.

B) Yes; the order is X-Y-Z.

C) Yes; the order is Z-X-Y.

D) No; based on this data alone, the order could be Z-Y-X or X-Y-Z.

E) No; based on this data alone, the order could be X-Z-Y or Z-X-Y.

Question 19 - 20. You have performed the following dihybrid cross in Drosophila using the black body color (b) and vestigial wing (vg) mutations. The b+ (grey body) and vg+ (normal wing) are dominant wild type alleles. These genes are autosomal.

Female â b+ vg+/b vg Ã male â b vg/b vg

Progeny:

Phenotype # of Progeny

Grey body normal wing 965

Black body vestigial wing 944

Grey body vestigial wing 208

Black body normal wing 195

19. Assuming linkage between black and vestigial, the estimated recombination frequency would be:

0.17

0.09

0.82

1.00

0.50

20. What key test could you use to determine whether the observed offspring frequencies deviate from those expected by chance alone?

A) Pascal's triangle

B) The product rule

C) The Chi-square (Ï²) test

D) The law of random assortment

E) The sum rule

21. In a genome wide association study (GWAS) designed to map the gene(s) that control height you divide subjects into a group of 1000 who are all more than seven feet tall and a control group of 1000 people of average height. You find the following associations between two genetic markers and the height trait:

		Tall group	Control group
Marker 1	Allele A	20%	50%
	Allele T	80%	50%

Marker 2	Allele G	15%	15%
	Allele C	85%	85%

What is your best guess for which marker is more closely linked to a gene that influences height?

A) Marker 1

B) Marker 2

22. Two pure breeding parents produce red and white flowers. They are crossed and the F1 produces pink flowers. When the F1 are selfed to produce the F2, nine distinct classes of pigmentation are present among F2 individuals. What is your best guess of the minimum number of genes that underlie flower pigmentation in this species?

A. 2

B. 3

C. 4

D. 5

E. 6

23. In a quantitative genetic experiment you identify two genes that confer bands of color on the back of a fly. At each gene, a dominant allele causes one band of color. If flies that are heterozygous at both loci are crossed, what ratio of offspring do you expect in each phenotype (i.e., number of color bands) class? (answer options are given from lowest to highest band number)

A) 1:1:1:1:1

B) 1:2:2:2:1

C) 1:4:6:4:1

D) 4:4:4:4:4