BSCI 1511 Study Guide - Midterm Guide: Tandem Repeat, Wild Type, Copy-Number Variation

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apricotminnow395

3 Apr 2017

School

Vanderbilt University

Department

Biological Sciences

Course

BSCI 1511

Professor

Abbot Patrick, Julian Hillyer

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Document Summary

It takes one generation of random mating to randomize alleles. Can have phenotypic diversity without any genetic diversity: environment can cause differences. Can have genetic diversity without phenotypic diversity: different genes producing same phenotype, e. g. Mendelian locus: 3 genotypes (aa, aa, aa) and only 2 phenotypes. Why genetic diversity does not equal phenotypic variation. If genes are more than 50 map units apart, their recombination frequency is 50: essentially on different chromosomes. Sickle cell, tay-sachs, achondroplasia, color-blindness, drosophilia: know what is dominant, what is recessive, red eyes dominant in fruit fly, white eyes recessive. Tay-sachs disease: wild type breaks down lipids, mutant does not breakdown lipids accumulation of lipids. What is the frequency of a sex-linked phenotype: for females, allele frequency = q, genotype = q2, for males = genotype frequency = allele frequency (b/c only 1 allele) Population genetics #6: q = 0. 08, p = 0. 92, 2pq = 2(0. 08)(0. 92)

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

In this study you will learn about a modification of Mendelian inheritance which is also an important source of genetic variation.

You are about to perform what looks like a classic dihybrid testcross ("A" is the gene for eye color and "B" is the gene for wing color.)

A allele for red eyes

a allele for white eyes

B allele for clear wings

b allele for spotted wings

(AaBb) X (aabb) Note the genotypes and phenotypes.

Question #1: Predict the expected phenotypic ratio of this P X P cross.

Observed results:

500 red eyes clear wings

500 white eyes spotted wings

Question #2: How do the expected results compare with the observed results?

Question #3: Considering Mendel's second law (regarding dihybrid crosses), suggest an explanation for the discrepancy between the observed results and those predicted.

Genes are located at specific positions on a chromosome. The distance between genes is measured in terms of "mapping units". In this exercise the gene for wing phenotype âBâ will be moved from its original position of zero to different positions on the chromosome. You will observe the effect of this gene movement.

First, gene "B" (wing color) is moved from position zero to position 10 on the same chromosome. Gene "A" (eye color) stays at position 0.

Observed Results: (position 10)

450 red eyes clear wings

50 red eyes spotted wings

50 white eyes clear wings

450 white eyes spotted wings

Question #4: How do these results differ from those predicted by Mendel's second law?

Question #5: From those in the first cross?

Question #6: Can you suggest a hypothesis to explain these new results?

Gene "B" (wing color) is now moved to position 20 on the same chromosome. Gene "A" (eye color) stays at position 0.

Observed Results: (position 20)

400 red eyes clear wings

100 red eyes spotted wings

100 white eyes clear wings

400 white eyes spotted wings

Question #7: How do the proportions of the phenotypic classes in this cross differ from the proportions you obtained in the earlier crosses?

The wing color gene "B" is now moved farther away (positions 30, 40, 50, and 60).

Observed Results: (position 30)

350 red eyes clear wings

150 red eyes spotted wings

150 white eyes clear wings

350 white eyes spotted wings

Observed Results: (position 40)

292 red eyes clear wings

220 red eyes spotted wings

195 white eyes clear wings

293 white eyes spotted wings

Observed Results: (position 50)

250 red eyes clear wings

250 red eyes spotted wings

250 white eyes clear wings

250 white eyes spotted wings

Observed Results: (position 60)

250 red eyes clear wings

250 red eyes spotted wings

250 white eyes clear wings

250 white eyes spotted wings

Question #8: Prepare a line graph plotting the % of recombinant offspring (Y-axis) vs. difference in position between the two genes (X-axis). (Hint: your graph should be one line with labeled points at map positions 0, 10, 20, 30, 40, 50, & 60).

Question #9: A genetic definition of "1 map unit" is the distance between 2 genes that gives 1% recombinants out of the total number of progeny. Using this definition, how many map units separate the "A" and "B" genes when the eye color gene "A" is at position 0 and the wing color gene "B" is at position 40? (Show how you determined the results.)

Question #10: How does the distance between two genes on the same chromosome affect the proportion of recombinants? Why?

Question #11: Maximum crossover frequency can be defined as the point at which no more recombination can be detected no matter how far apart the genes are located. Do your results indicate that you are approaching the maximum crossover frequency?

Question #12: Show (using a Punnett Square) the expected phenotypic rations if the two genes were located on different chromosomes.

Question #13: Which of your simulated cross or crosses do the data from the Punnett Square most closely fit? Why?

Question #14: A wild-type fly (heterozygous for gray body color and normal wings - b⁺ vg⁺/b vg) was mated to a black fly with vestigial wings (b vg/b vg). The F1 had the following phenotypic distribution: wild type, 850; black body - vestigial wings, 899; black body - normal wings, 79; gray body - vestigial wings, 85. What is the recombination frequency (RF) between these genes for body color and wing type? (show work)

Question #15: A cross produced 915 offspring with normal pigment and 310 with albinism. Conclusion?

a. One of the parents was homozygous for albinism.

b. Both parents were heterozygous.

c. One parent was homozygous for normal pigmentation.

d. Both parents were albinos. e. 605 albino zygotes must have failed to develop.

NOTE: Please show all work to support your answers.

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BSCI 1511 Study Guide - Midterm Guide: Tandem Repeat, Wild Type, Copy-Number Variation

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