BIO 352 Lecture Notes - Lecture 5: Chromosome, Gene Mapping, Mendelian Inheritance

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Gene Linkage: Two genes are localized on a single chromosome, and will therefore be

inherited as a single unit, therefore no independent assortment. Can break linkage

through recombination events

Gene mapping; two linked genes on same chromosome, if they are really close the

frequency of recombination between the two linked genes would be very low. However,

if they are linked but far apart, the frequency of recombination between the two will be

higher.

oRecombination frequency will tell you about the distance of gene linkage

oUsing the recombination frequencies, maps can be created

Linkage

oSame genotype, different ways they are linked relative to each other, (cis or

trans)

Genetically significant crossing-over only occurs between non-sister chromatids

Through recombination you go from cis → trans

Correlation between linked genes on chromosome and recombination frequency

Meiotic Recombination

oIndependent Assortment

AaBb

¼ AB

¼ Ab

¼ aB

¼ ab

oAbsolute Linkage (no recombination)

Cis AB/ ab

½ AB

½ ab

Trans Ab/aB

½ Ab

½ aB

oLinkage with recombination

Trans AB/ ab

X = recombination frequency

X = 0 no recombination

½ - x/2 AB

x/2 Ab

x/2 aB

½ - x/2 ab

Same gametic types as independent assortment but present in different

proportions

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Document Summary

Gene linkage: two genes are localized on a single chromosome, and will therefore be inherited as a single unit, therefore no independent assortment. Gene mapping; two linked genes on same chromosome, if they are really close the frequency of recombination between the two linked genes would be very low. However, if they are linked but far apart, the frequency of recombination between the two will be higher: recombination frequency will tell you about the distance of gene linkage, using the recombination frequencies, maps can be created. Linkage o: same genotype, different ways they are linked relative to each other, (cis or trans) Through recombination you go from cis trans. Genetically significant crossing-over only occurs between non-sister chromatids. Correlation between linked genes on chromosome and recombination frequency. Ab: absolute linkage (no recombination) X/2 ab x/2 ab x/2 ab. Same gametic types as independent assortment but present in different proportions.

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

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%

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

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BIO 352 Lecture Notes - Lecture 5: Chromosome, Gene Mapping, Mendelian Inheritance

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