BIOL239 Lecture Notes - Lecture 8: Meiosis, Chromosome, Genotype

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amethystdonkey419

14 May 2018

School

University of Waterloo

Department

Biology

Course

BIOL239

Professor

Christine Dupont

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

Set 8: Linkage and Recombination

●Some genes on the same chromosome assort together more often than not

●In dihybrid crosses, departures from a 1:1:1:1 genotypic ratio of F1 gametes indicates that

the two genes are on the same chromosome – not completely independent of each other;

can see this in F2 offspring

oThey are in close proximity to one another and therefore tend to assort together –

genetically linked

oSome genes are more closely linked than others

●Parental class – phenotype of P generation (not parents of F2)

●When genes are linked, parental combinations (more P features expressed) outnumber

recombinant types Drosophila



– model for genetics

o in testcross…

o 82% similar to either one of the parents (ie. no crossover event)

o 18% result from recombinations (crossing over; nonparental; random and rare)

o therefore 9:3:3:1 ratio may not show; still we see dominance of parental

phenotypes

▪ knowing that some genes travel closely together

●~75% of known human disease genes have a similar gene match in Drosophila



genome

●~50% of fly protein sequences have a similar protein match in mammals

●Used in research on neurodegenerative diseases, e.g. Parkinson’s, Huntington,

Alzheimer’s; research on processes of aging, diabetes, drug abuse

●Only concerned with phenotype when classing

●Recombination between non-sister chromatids during meiosis (pachynema) gives rise to

recombinant phenotypes (trading of alleles)

●The number of individuals in the recombinant phenotype classes varies with how closely

the genes are linked

●The number in the recombinant class will always be less than the number in the parental

class if the genes are linked

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Document Summary

Some genes on the same chromosome assort together more often than not. Parental class phenotype of p generation (not parents of f 2 ) When genes are linked, parental combinations (more p features expressed) outnumber recombinant types drosophila model for genetics o o o o in testcross . 82% similar to either one of the parents (ie. no crossover event) 18% result from recombinations (crossing over; nonparental; random and rare) therefore 9:3:3:1 ratio may not show; still we see dominance of parental phenotypes knowing that some genes travel closely together. ~75% of known human disease genes have a similar gene match in drosophila genome. ~50% of fly protein sequences have a similar protein match in mammals. Used in research on neurodegenerative diseases, e. g. parkinson"s, huntington, Alzheimer"s; research on processes of aging, diabetes, drug abuse. Recombination between non-sister chromatids during meiosis (pachynema) gives rise to recombinant phenotypes (trading of alleles)

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

aquamarinemoose208

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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BIOL239 Lecture Notes - Lecture 8: Meiosis, Chromosome, Genotype

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