FALL 2005- WEEK 15 SOLUTIONS

Chapter Seven: Linkage, Recombination, and Eukaryotic Gene Mapping

 

COMPREHENSION QUESTIONS

 

*1.    What does the term recombination mean? What are two causes of recombination?

Recombination means that meiosis generates gametes with different allelic combinations than the original gametes the organism inherited. If the organism was created by the fusion of an egg bearing AB and a sperm bearing ab, recombination generates gametes that are Ab and aB. Recombination may be caused by loci on different chromosomes that sort independently or by a physical crossing over between two loci on the same chromosome, with breakage and exchange of strands of homologous chromosomes paired in meiotic prophase I.

 

*2.    In a testcross for two genes, what types of gametes are produced with (a) complete linkage, (b) independent assortment, and (c) incomplete linkage?

(a) Complete linkage of two genes means that only nonrecombinant gametes will be produced; the recombination frequency is zero.

(b) Independent assortment of two genes will result in 50% of the gametes being recombinant and 50% being nonrecombinant, as would be observed for genes on two different chromosomes. Independent assortment may also be observed for genes on the same chromosome, if they are far enough apart that one or more crossovers occur between them in every meiosis.

(c) Incomplete linkage means that greater than 50% of the gametes produced are nonrecombinant and less than 50% of the gametes are recombinant; the recombination frequency is greater than 0 and less than 50%.

 

3.      What effect does crossing over have on linkage?

Crossing over generates recombination between genes located on the same chromosome, and thus renders linkage incomplete.

 

4.      Why is the frequency of recombinant gametes always half the frequency of crossing over?

Crossing over occurs at the four-strand stage, when two homologous chromosomes, each consisting of a pair of sister chromatids, are paired. Each crossover involves just two of the four strands and generates two recombinant strands. The remaining two strands that were not involved in the crossover generate two nonrecombinant strands. Therefore, the frequency of recombinant gametes is always half the frequency of crossovers.

 

 

 

 

 

 

6.      How does one test to see if two genes are linked?

One first obtains individuals that are heterozygous for both genes. This may be achieved by crossing an individual homozygous dominant for both genes to one homozygous recessive for both genes, resulting in a heterozygote with genes in coupling configuration. Alternatively, an individual that is homozygous recessive for one gene may be crossed to an individual homozygous recessive for the second gene, resulting in a heterozygote with genes in repulsion. Then the heterozygote is mated to a homozygous recessive tester and the progeny of each phenotypic class are tallied. If the proportion of recombinant progeny is far less than 50%, the genes are linked. If the results are not so clear-cut, then they may be tested by chi-square, first for equal segregation at each locus, then for independent assortment of the two loci. Significant deviation from results expected for independent assortment indicates linkage of the two genes.

 

7.      What is the difference between a genetic map and a physical map?          

A genetic map gives the order of genes and relative distance between them based on recombination frequencies observed in genetic crosses. A physical map locates genes on the actual chromosome or DNA sequence, and thus represents the physical distance between genes.

 

*8.    Why do calculated recombination frequencies between pairs of loci that are located relatively far apart underestimate the true genetic distances between loci?

The further apart two loci are, the more likely it is to get double crossovers between them. Unless there are marker genes between the loci, such double crossovers will be undetected because double crossovers give the same phenotypes as nonrecombinants. The calculated recombination frequency will underestimate the true crossover frequency because the double crossover progeny are not counted as recombinants.

 

9.      Explain how one can determine which of three linked loci is the middle locus from the progeny of a three-point testcross.

Double crossovers always result in switching the middle gene with respect to the two nonrecombinant chromosomes. Hence, one can compare the two double crossover phenotypes with the two nonrecombinant phenotypes and see which gene is reversed. In the diagram on the facing page we see that the coupling relationship of the middle gene is flipped in the double crossovers with respect to the genes on either side. So whichever gene on the double crossover can be altered to make the double crossover resemble a nonrecombinant chromosome is the middle gene. If we take either of the double crossover products l M r or L m R, changing the M gene will make it resemble a nonrecombinant.

                 l        m         r                     l         M         r

                            

               L       M        R                     L         m         R

 

APPLICATION QUESTIONS AND PROBLEMS

 

*13.  In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross).

(a) What will be the results of the testcross if the loci that control banding and color are linked with no crossing over?

      With absolute linkage, there will be no recombinant progeny. The F1 inherited banded and yellow alleles (BBCY) together on one chromosome from the banded yellow parent and unbanded and brown alleles (BOCBw) together on the homologous chromosome from the unbanded brown parent. Without recombination, all the F1 gametes will contain only these two allelic combinations, in equal proportions. Therefore, the F2 testcross progeny will be ½ banded, yellow and ½ unbanded, brown.

(b) What will be the results of the testcross if the loci assort independently?

      With independent assortment, the progeny will be:

      ¼ banded, yellow

      ¼ banded, brown

      ¼ unbanded, yellow

      ¼ unbanded, brown

(c) What will be the results of the testcross if the loci are linked and 20 map units apart?

      The recombination frequency is 20%, so each of the two classes of recombinant progeny must be 10%. The recombinants are banded, brown and unbanded, yellow. The two classes of nonrecombinants are 80% of the progeny, so each must be 40%. The nonrecombinants are banded, yellow and unbanded, brown. In summary:

      40% banded, yellow

      40% unbanded, brown

      10% banded, brown

            10% unbanded, yellow

 

*14.  In silkmoths (Bombyx mori), red eyes (re) and white-banded wing (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have normal eyes and normal wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are:

      wild-type eyes, wild-type wings                                 418

      red eyes, wild-type wings                                            19

      wild-type eyes, white-banded wings                            16

      red eyes, white-banded wings                                   426

 

(a) What phenotypic proportions would be expected if the genes for red eyes and white-banded wings were located on different chromosomes?

¼ wild-type eyes, wild-type wings

¼ red eyes, wild-type wings

¼ wild-type eyes, white-banded wings

¼ red eyes, white-banded wings

(b) What is the genetic distance between the genes for red eyes and white-banded wings?

       The F1 heterozygote inherited a chromosome with alleles for red eyes and white-banded wings (re wb) from one parent and a chromosome with alleles for wild-type eyes and wild-type wings (re+ wb+) from the other parent. These are therefore the phenotypes of the nonrecombinant progeny, present in the highest numbers. The recombinants are the 19 with red eyes, wild-type wings and 16 with wild-type eyes, white-banded wings.

       RF = recombinants/total progeny × 100% = (19 + 16)/879 × 100% = 4.0%

       The distance between the genes is 4 map units.

 

*24.(a and b only)        Waxy endosperm (wx), shrunken endosperm (sh), and yellow seedling (v) are encoded by three recessive genes in corn that are linked on chromosome 5. A corn plant homozygous for all three recessive alleles is crossed with a plant homozygous for all the dominant alleles. The resulting F1 are then crossed with a plant homozygous for the recessive genes in a three-point testcross. The progeny of the testcross are given below:

                wx     sh      V          87

                Wx    Sh     v           94

                Wx    Sh     V          3479

                wx     sh      v           3478

                Wx    sh      V          1515

                wx     Sh     v           1531

                wx     Sh     V          292

                Wx    sh      v           280

                total                      10,756

 

(a) Determine order of these genes on the chromosome.

      The nonrecombinants are Wx Sh V and wx sh v.

      The double crossovers are wx sh V and Wx Sh v.

      Comparing the two, we see that they differ only at the v locus, so v must be the middle gene.

(b) Calculate the map distances between the genes.

      Wx-V distance—recombinants are wx V and Wx v:

      RF = (292+280+87+94)/10,756 = 753/10,756 = .07 = 7% or 7 m.u.

      Sh-V distance—recombinants are sh V and Sh v:

      RF = (1515+1531+87+94)/10,756 = 3227/10,756 = 30 = 30% or 30 m.u.

      The Wx-Sh distance is the sum of these two distances: 7 + 30 = 37 m.u.l.

 

25.(a and b only) Fine spines (s), smooth fruit (tu), and uniform fruit color (u) are three recessive traits in cucumbers whose genes are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross and the progeny at the top of the following page are produced from this testcross.

               

                S        U      Tu          2

                s        u       Tu          70

                S        u       Tu          21

                s        u       tu           4

                S        U      tu           82

                s        U      tu           21

                s        U      Tu          13

                S        u       tu           17__

                total                          230

 

(a) Determine the order of these genes on the chromosome.

Nonrecombinants are s u Tu and S U tu.

Double crossovers are s u tu and S U Tu.

Because Tu differs between the nonrecombinants and the double crossovers, Tu is the middle gene.

(b) Calculate the map distances between the genes.

      S-Tu distance: recombinants are S Tu and s tu.

      RF = (2 + 4 + 21 + 21)/230 = 48/230 = 21% or 21 m.u.

      U-Tu distance: recombinants are u tu and U Tu.

      RF = (2 + 4 + 13 + 17)/230 = 36/230 = 16% or 16 m.u.