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HMB265H1 Lecture 9: HMB265 Lecture 9 READINGS - Linkage Mapping

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rosebadger889
6 Oct 2017
School
UTSG
Department
Human Biology
Course
HMB265H1
Professor
Maria Papaconstantinou

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HMB265H1 Full Course Notes
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Document Summary

Hmb265 lecture 9 readings - linkage mapping (pg. Recombination frequency reflects the distance between two genes. Different gene pairs = different linkage frequencies b/c gene pairs arranged in a line along a chromosome. Closer together 2 genes on a chromosome = less chance of being separated by an event that cuts and recombines the line of genes. Recombination frequency (rf): percent of total progeny that were recombinant types. Can use rf to gauge physical distance separating any two genes on same chromosome. Arbitrarily defined one rf percentage point as unit of measure along a chromosome. Can also use centimorgan (cm) or use map unit (m. u. ) 1% rf = 1 cm = 1 m. u. Ex: x-linked genes for eye colour + body colour recombine in 1. 1% of f2 progeny = 1. 1 m. u. Fraction of recombinant gametes = measure of distance between 2 genes: If genes a and b close together on chromosome, most likely no crossovers between the genes.

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Related Questions

ANSWER ALL the PARTS you can

PART1 Alleles are considered linked if recombination is less than 50% in:

a) a Px P Fl cross

b) an F1 x F1F2 cross

c) an F2 x F2F3 cross

d) an Fl x homozygous recessive cross

e) any F2 progeny x homozygous recessive cross 1.

PART2 If two loci A/a B/b are 9 map units apart, and an Ab/aB individual is testcrossed,

a) There will be 9% AB individuals

b)There will be 91% Ab individuals

c) There will be 4.5% ab individuals

d) There will be 91% recombination individuals

If two genes a and b show 50% recombination genes a and c show 35% recombination, and genes b and show 32% recombination, then

a) Genes a and b are on the same chromosome, and 50 m.u. apart

b) Genes a and b are on different chromosomes

c)Genes a and b are on the same chromosome, and 67 map units apart

d) One cannot tell if genes a and b are on the same or different chromosomes from given data.

PART4 If two genes a and b are 8.3 m.u. a part, one expects to find

a) 8.3% recombinant gametes from a double heterozygous parent

b) 91.7% recombinant gametes from a double heterozygous parent

c) A chiasma between the a and b loci in 8.3% of the meiosis

d) A chiasma between the a and b loci in 91.7% of the meiosis.

PART5 Map distances are most accurate when genes are closely linked because as genes become further apart,

a) They always show independent assortment

b)The chance of multiple crossovers increases

c) They no longer recombine

d) There is chiasma interference

PART6 In a three-point testcross, the most frequent classes of progeny were ABc and abC. The less frequent classes of progeny were AbC and aBc. Which gene is in the middle?

a) Aa

b) B/b

c)C/c

d) It is not possible to tell from just these data

PART7 In a three-point testcross with 100 progeny, there were 16 single crossovers between genes A and B, 29 single crossovers between genes B and C, and 4 double crossovers. The correct gene order was determined to be A-B-C. What is the map distance between A and B?

a) 20%

b) 0.2 m.u

c) 20 m.u.

d) 16 m.u.

e) 29 m.u.

f) 16%

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 “B” 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.