Two mutations, a and b, are found to lie 5 map units apart. What will be the frequency of recombinant chromosomes generated from a cis heterozygote in C. elegans?
The Mississippi Bippy is a hypothetical animal favored by Columbia geneticists. In this animal amiable (A) is dominant to nasty (a), lazy (L) is dominant to active (l), and crazy (C) is dominant to sane (c). An amiable, lazy, but sane female is crossed with a nasty, active, but crazy male and all of the F1 progeny are found to be amiable, lazy, and crazy. The testcross of an F1 female with a nasty, active, and sane male gave the following results:
Phenotype |
# |
amiable, lazy, crazy |
56 |
amiable, lazy, sane |
413 |
amiable, active, crazy |
1 |
amiable, active, sane |
22 |
nasty, lazy, crazy |
24 |
nasty, lazy, sane |
3 |
nasty, active, crazy |
421 |
nasty, active, sane |
60 |
Draw out the crosses (i.e. indicate the genotypes of the parents).
What are the genotypes of the progeny.
Construct a genetic map showing the map order and the distances between the genes.
What is genetic interference? Do the data show evidence of interference? If so, how much is there?
Mapping
How many map units must separate two genes on a chromosome before it is impossible to distinguish linkage from nonlinkage? (Hint: you might try some representative recombination frequencies to test out your ideas.)
Some chromosomes in Drosophila and yeast have genes that are separated by over 100 map units. How can having chromosomes with this many map units be possible given your answer to part b?
You are asked to determine the map distance between two recessive mutations (a and b) by examining the progeny of a cross between a cis heterozygote female and a trans heterozygote male (both male and female germlines exhibit recombination).
What are genotypes (appropriately labeled) of the parents.
What would be the parental phenotype(s) and recombinant phenotype(s) among the progeny from this cross?
Give a formula to determine the map distance (p) between these two mutations.
Would determining p from this cross be harder or easier than determining from a cross of two trans heterozygotes? Why?
One method of mapping the position of gene c relative to two genes, a and b, on the same chromosome in Drosophila that is usually described in textbooks is to construct a c/ab heterozygote, do a test cross with an abc/abc animal and look at the rarest class (double recombinants). However, if c is a newly identified gene whose position you are trying to map, you probably won't have abc/abc animals. Another method is to mate the c/ab heterozygote to an ab/ab animal and then mate progeny with the A or B phenotype with c/c animals.
For the following situations, give the expected genotypes of all of the A (i.e. A non B) and B (i.e B non A) progeny from the mating of c/ab with ab/ab. Assume that all three genes are fairly close and that b is to the right of a. (Hint: draw the chromosomes of the heterozygote and determine where relevant crossovers will occur.)
c is to the right of a and b.
c is to the left of a and b.
c is between a and b.
What progeny will result when the A and B animals in (a) are mated to c/c animals?
Use the information you have obtained in (a) and (b) to formulate a rule for mapping the position of c relative to a and b.
Why were only A and B progeny from (a) used in this method? In other words, what is wrong with all the other possible progeny from the cross?
Such a mapping experiment was done with the following results (many other progeny were ignored). If a and b are 2.4 map units apart, calculate the position of c relative to a and b.
15 A animals produced C progeny
7 A animals did not produce C animals
4 B animals produced C progeny
18 B animals did not produce C animals
Now,
Redo part (a)(i) for the situation where a and b are fairly close and c is very far to the right of both genes. What problem does this result present?
How could this problem be resolved?
Often chromosomes need to be generated that contain mutations in two genes, i.e. in the cis configuration.
One method would be to construct a trans heterozygote, and look for recombinants. Suppose that two genes, a and b, are 3 map units apart. Trans heterozygotes (a+/+b) are mated to each other.
What proportion of the progeny from the cross will have the double mutant phenotype (both the a and b mutations are recessive)?
What proportion of the progeny contain at least one chromosome with both mutations in cis?
Using the above information and using a self-fertilizing species, devise a method to generate ab/ab homozygotes.
A second method is to take advantage of a three-factor cross. Using your results from problem 5, devise a scheme to generate an ab chromosome if you have strains that are ac/ac and b/b. Be sure to indicate what criteria you would look for in choosing an appropriate ac/ac strain for this construction.
Four recessive mutations that result in uncoordinated movement were identified in a mutagenesis of C. elegans. Animals from four true-breeding lines of each mutation were mated to each other. The F1 cross progeny from these matings were allowed to selfcross with the following results:
Cross F1 F2
1 X 2 all uncoordinated all uncoordinated
1 X 3 all coordinated 9/16 coordinated, 7/16 uncoordinated
1 X 4 all coordinated 1/2 coordinated, 1/2 uncoordinated
Explain these results (use diagrams wherever possible).
What F1 and F2 cross progeny would you predict from crosses of (i) 2 X 3 and (ii) 2 X 4?
You are asked to determine the map distance (p) between a dominant mutation (a) and a recessive mutation (b) by examining the progeny of a cross between a cis heterozygous female and a trans heterozygous male (recombination occurs equally in more sexes in this species).
Draw the genotype (appropriately labeled) of the parents.
What would be the parental phenotype(s) and recombinant phenotype(s) among the progeny of this cross?
Give a formula to determine the map distance between these two mutations (assume that p is small).
Would determining p from this cross be harder, easier, or about the same than determining it from a cross of two cis heterozygotes? Explain your answer.
In Drosophila a female heterozygous for recessive mutations in three genes (sc, ec, and vg) is mated to a male that is homozygous for all three mutations. The following progeny were found.
sc ec vg 233
+ + + 239
sc ec + 241
+ + vg 231
sc + vg 12
+ ec + 14
sc + + 14
+ ec vg 16
+ 1000
What is the genotype of the female heterozygote?
Which of the genes, if any, are linked?
Draw a genetic map showing the placement of all three gene (be sure to indicate map distances if they can be calculated.
You are forced to work in the Chalfie lab mapping the position of three genes (x, y, and z) in C. elegans. Both genes appear to be near other two genes (a and b) that are known to be 3.0 map units apart. Mutations x and y are mapped relative to a and b by making heterozygotes x/ab and y/ab, collecting A and B progeny, and looking at the progeny of these animals for any that show the X (or Y) phenotype. The results of two crosses are given below. By examining the data, give as accurate description of the map position of x and y relative to a and b as you can. Explain your answers.
For x/ab:
All of the 25 A progeny had some progeny that also had the X phenotype.
None of the 19 B progeny had progeny with the X phenotype.
For y/ab:
16 of 21 A progeny had some progeny that also had the Y phenotype.
2 of 14 B progeny had progeny with the Y phenotype.
Explain how the trans configuration is better than the cis configuration to determine the 2-factor map distance between two mutations in C. elegans if one mutation is dominant and the other is recessive.
C. elegans hermaphrodites have two X chromosomes, males have only one X. Males arise infrequently by nondisjunction. Several years ago a translocation joining almost all of the X chromosome to almost all of chromosome I (we'll call it T) was identified. Animals that are homozygous for T do not survive (some of the missing genes on both chromosomes are essential), but animals with the genotype I/T/X (i.e. one copy of each I, T, and X) survive.
What will be the products of meiosis in such a heterozygous strain?
Demonstrate that I/T/X animals will produce the same types of viable progeny as a strain homozygous for a mutation causing X nondisjunction.
Suppose the I/T/X animal had a recessive egl mutation on chromosome I and a recessive unc mutation on the X chromosome. The T chromosome has wild-type alleles for both genes. (i) What is the phenotype of this animal? (ii) How would the Unc phenotype segregate compared with the Egl phenotype? (iii) How does the presence of a translocation affect linkage?
Explain the following:
How translocations lead to apparent (pseudo) linkage of two normally unlinked mutations.
How recombination would be reduced in the region near a translocation breakpoint in a translocation heterozygote (think about the structure of the meiotic structure that is produced).
How inversions seem to prevent recombination.
How different translocations can be used to generate strains with duplications and/or deletions.
Distinguish briefly between each set of the following pairs
Linkage and pseudolinkage
translocation and inversion
Describe genetic (not molecular) experiments to determine which chromosomes are joined in a translocation in C. elegans (or Drosophila).
Answer TRUE or FALSE. If the statement is TRUE, explain or give a supporting example. If the statement is FALSE, either correct the statement or give a counterexample.
Recombinant gametes are not found in strains with paracentric inversions because pairing is prevented during meiosis I.
A translocation joining two chromosomes (e.g., I and III) does not affect the independent assortment of mutations on nontranslocation homologues of these chromosomes.