Linkage and Chromosome Mapping Practice Problems (page 3)

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Updated on Apr 25, 2014

Practice 7

Suppose we are given a strain of Drosophila exhibiting an unknown abnormal genetic trait (mutation).We mate the mutant females to males from a balanced lethal strain (Cy Pm+ /Cy+ Pm, DSb+ /D+ Sb) where curly wings (Cy) and plum eye (Pm) are on chromosome 2 and dichaete wing (D) and stubble bristles (Sb) are on chromosome 3. Homozygosity for either curly, plum, dichaete, or stubble is lethal. The trait does not appear in the F1. The F1 males with curly wings and stubby bristles are then backcrossed to the original mutant females. In the progeny the mutation appears in equal association with curly and stubble. Drosophila melanogaster has a haploid number of 4 including an X, 2, 3, and 4 chromosome.     (a)  Determine whether the mutation is a dominant or a recessive.     (b)  To which linkage group (on which chromosome) does the mutation belong?

Solution 7

(a)  If the mutation were a dominant (let us designate it M), then each member of the strain (pure line) would be of genotype MM. Since the trait does not appear in our balanced lethal stock, they must be homozygous recessive (M+M+). Crosses between these two lines would be expected to produce only heterozygous genotypes (M+M) and would be phenotypically of the mutant type. But since the mutant type did not appear in the F1, the mutation must be a recessive (now properly redesignated m). The dominant wild-type allele may now be designated m+.

(b)  Let us assume that this is a sex-linked recessive mutation. The F1 males receive their single X chromosome from the mutant female (mm). Therefore, all males of the F1 should exhibit the mutant trait because males would be hemizygous for all sex-linked genes (mY). Since the mutant type did not appear in the F1, our recessive mutation could not be sex-linked.

Let us assume that our recessive mutation is on the second chromosome. The curly, stubble F1 males carry the recessive in the heterozygous condition (Cy m+ /Cy+ m, Sb/Sb+). Notice that we omit the designation of loci with which we are not concerned. When these carrier males are then backcrossed to the original mutant females (Cy+ m/Cy+ m, Sb+ /Sb+), the F2 expectations are as follows:

Note that the mutant cannot appear with curly. Therefore, our recessive mutation is not on chromosome 2.

Let us then assume that our mutant gene is on the third chromosome. When F1 carrier males (Cy/Cy+, Sb m+ /Sb+ m) are backcrossed to the original mutant females (Cy+ /Cy+, Sb+ m/Sb+ m), the F2 expectations are as follows:

Note that the mutant cannot appear with stubble. Hence, our recessive mutation is not on chromosome 3.

If the mutant is not sex-linked, not on 2 nor on 3, then obviously it must be on the fourth chromosome. Let us prove this. When F1 carrier males (Cy/Cy+, Sb/Sb+, m+/m) are backcrossed to the original mutant females (Cy+/Cy+, Sb+/Sb+, m/m), the F2 expectations are as follows:

Solved Problems

Note that our recessive mutant occurs in equal association with curly and stubble, which satisfies the conditions of the problem. We conclude that this mutation is on the fourth chromosome.

Practice 8

A strain of Neurospora requiring methionine (m) was crossed to a wild-type (m+) strain with the results shown below. How far is this gene from its centromere?

Solved Problems

Solution 8

Noncrossover asci are those that appear with the greatest frequency = 40 + 36 = 76 out of 100 total. The other 24/100 are crossover types. While 24% of the asci are crossover types, only half of the spores in these asci are recombinant. Therefore, the distance from the gene to the centromere is 12 map units. The origin of the crossover-type asci is as follows:

Solved Problems

Practice 9

Two strains of Neurospora, one mutant for gene a, the other mutant for gene b, are crossed. Results are shown below. Determine the linkage relationships between these two genes.

Solved Problems

Solution 9

Pattern 1 represents the noncrossover types showing first-division segregation (4 : 4) for both a and b. Pattern 2 shows second-division segregation (2 : 2 : 2 : 2) for a, but first-division segregation for b. Genes that show high frequency of second-division segregation are usually further from the centromere than genes with low frequency of second-division segregation. Judging by the relatively high frequency of pattern 2 these are probably single crossovers, and we suspect that a is more distal from its centromere than b.

Solved Problems

Pattern 3, indicating first-division segregation for a and second-division segregation for b, cannot be generated by a single crossover if a and b are linked as shown above, but requires a double crossover.

Solved Problems

Furthermore, pattern 4 could be produced from the linkage relationships as assumed above by a single crossover in region I.

Solved Problems

Double crossovers (DCO) are expected to be much less frequent than single crossovers (SCO). Under the above assumptions, double-crossover pattern 3 is more frequent than one of the single-crossover patterns 4. This does not make sense, and thus our assumption must be wrong. The locus of a must be further from the centromere than b, but it need not be on the same side of the centromere with b. Let us place a on the other side of the centromere.

Solved Problems

Now a single crossover in region I produces pattern 2, a single crossover in region II produces pattern 3, and a two strand double crossover (I, II) produces pattern 4. The percentage of asci are numerically acceptable under this assumption.

The distance a-centromere =1/2(SCOI + DCO) = 2(14 + 1) = 7.5 map units

The distance centromere-b = 1/2(SCOII + DCO) = 2(6 + 1) = 3.5map units

Practice 10

The cross (abc) × (+ + +) is made in an ascomycete with unordered tetrads. From the analysis of 100 asci, determine the linkage relationships between these three loci as completely as the data allow.

  1. 40 (abc)(abc)(+ + +)(+ + +)
  2. 42 (ab+)(ab+)(+ + c)(+ + c)
  3. 10 (a + c)(+ + c)(ab +)(+ b +)
  4. 8(a + +)(+ + +)(abc)(+ bc)

Solution 10

Pattern 1 is parental ditype (PD) for ab, ac, and bc. Pattern 2 is PD for ab, nonparental ditype (NPD) for ac and bc. Pattern 3 is tetratype (TT) for ab and ac, NPD for bc. Pattern 4, is TT for ab and ac, PD for bc. For each pair of markers the relative frequencies of each type of tetrad are as follows:

Solved Problems

For the ab pair, PDs are not equivalent with NPDs. Thus, a and b must be linked. For pairs ac and bc, PDs are roughly equivalent with NPDs. Thus, c must be assorting independently on another chromosome. The recombination frequency between a and b is

The recombination frequency in percent is equal to 9 map units. Single crossovers between either b and its centromere or c and its centromere, or both, would produce TTs for bc. Since none occurred we can assume that the loci of b and c are both very near their respective centromeres.

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