BI226 Chapter 5: Gene Linkage

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11 Feb 2017

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Chapter 5 - linkage, recombination, and the mapping of genes on chromosomes. Linkage, recombination, and mapping of genes on chromosomes. Genes linked together on the same chromosome usually assort together. Detecting linkage by analyzing the progeny of dihybrid crosses: x linked genes. Synthetic genes - located on the same chromosome. Two x linked genes in drosophila with recessive alleles. W + (red eyes) and w (white eyes) Y + (brown body) and y (yellow body) F1 males get their only x chromosomes from their mothers. Compare allele configurations in f 2 to p generation. Deviation from 1:1:1:1 segregation in f 2 indicates the genes are linked. Note that in this cross involving x linked genes, only male progeny were counted the f. Designation of parental and recombinant relate to past history. Note that the parental configurations in these two crosses are opposite each other. Detect linkage by generating a double heterozygote and crossing to homozygous recessive (testcross)

Related Questions

In a series of crosses to look at the relationship of different genes with each other, the following crosses were performed. The numbers of offspring obtained from each cross are shown.

Cross offspring linkage?

AaBb x aabb 20 AaBb, 15 Aabb, 18 aaBb, 17aabb

CcDd x ccdd 54 CcDd 49 ccdd, 98 Ccdd, 101 ccDd

EeFf x eeff 43 EeFf 7 Eeff 42 eeff 5 eeFf

a) For each one, note whether the genes are linked, and if so how closely (in mu); and whether the alleles are present in coupling or repulsion in the heterozygous parent.

b) Which genes are most tightly linked (closest together on the same chromosome)?

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%

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

BI226 Chapter 5: Gene Linkage

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