Chapter 5: Chromosome Mapping in Eukaryotes Flashcards
TRUE or FALSE?
Two genes that are separated by 10 map units show a recombination percentage of 10%.
True
One map unit is equal to 1% recombination between two genes; 10 map units would be equal to 10% recombination between the genes.
Genes X, Y, and Z are linked. Crossover gametes between genes X and Y are observed with a frequency of 25%, and crossover gametes between genes Y and Z are observed with a frequency of 5%. What is the expected frequency of double crossover gametes among these genes?
1.25%
The probability of a double crossover is the product of the probabilities of the single crossovers: 0.25 x 0.05 = 0.0125, or 1.25%.
Assume that the genes from the previous example are located along the chromosome in the order X, Y, and Z. What is the probability of recombination between genes X and Z?
30%
Recombination frequencies between linked genes along a chromosome are additive, so the recombination frequency between genes X and Z is 25 + 5 = 30.
The presence of wild-type F2 progeny indicates a dihybrid cross. The converse is also true: If a cross fails to produce wild-type progeny in the F2, it is a monohybrid cross, and the two lines are mutant in the same locus.
Put another way, there is no opportunity for independent assortment to produce a wild-type genotype in the F2 because the two lines fail to complement each other.
TRUE or FALSE?
Crossing over during prophase I of meiosis occurs between alleles on sister chromatids
FALSE
Crossing over during meiosis occurs between alleles on nonsister chromatids.
Meiosis Facts
Complete linkage results in the formation of only parental gametes.
Parental gametes contain the same combinations of linked genes as found in the parent cell.
Recombinant gametes contain combinations of alleles not found in the parent cell.
If crossing over occurs, half of the gametes formed are parental and the other half are recombinant.
If two genes on the same chromosome exhibit complete linkage, what is the expected F2 phenotypic ratio from a selfed heterozygote with the genotype a+b+ ⁄⁄ ab?
3:1
Each parent produces two types of gametes, a+b+ and ab, giving the simple Mendelian ratio of 3 a+b+ : 1 ab.
Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci mentioned above?
48 primary oocytes
TRUE or FALSE?
In a gene mapping cross, the term Frac NCO refers to the fraction of gametes that have the same genotypes as the parental gametes.
TRUE
Frac NCO refers to the fraction of gametes that have not undergone crossing over and thus their genotypes reflect those of the parental gametes.
the following statements about three linked genes that are spaced very close together along a chromosome are most likely to be true
Interference will be a significant factor in the number of crossovers observed.
Interference effects are more likely when crossovers are confined to a small region.
For linked genes A, B, and C, the fraction of single crossovers for genes A and B is 0.1, the fraction of single crossovers for genes B and C is 0.3, and the fraction of double crossovers is 0.03. Which of the following statements is true?
The distance between A and B is less than the distance between B and C.
The frequency of crossover events decreases as the distance between genes decreases.
In this case, the F1 is a dihybrid with AB on one chromosome and ab on the other chromosome. You can tell this by looking at the parents. The F1 hybrid got AB from one parent and ab from the other parent (AB/ab). A dihybrid that is AB/ab can produce four possible “output” gametes through meiosis. Output gametes that match one of the input gametes are nonrecombinant (AB or ab). Output gametes with new combinations of alleles are recombinant (Ab or aB).
Its important to realize that its also possible to have a dihybrid who is Ab/aB instead. In this case, one parent was AAbb and the other parent aaBB. A dihybrid who is Ab/aB can also produce four possible “output” gametes through meiosis, but in this case the nonrecombinant gametes are Ab or aB and the recombinant gametes are AB or ab. In Part C, be sure to check the parents of the F1 trihybrid to determine their configuration of alleles before answering the question. Which gametes are nonrecombinant versus recombinant depends how the alleles are organized in the F1 trihybrid.
In a testcross with a homozygous recessive line (tester), the tester will contribute only recessive alleles that do not affect the phenotypes of the testcross progeny. Therefore, such testcrosses allow you to determine the haploid genotype of the gametes produced by the F1 dihybrid.
Recombination occurs when combinations of alleles not found in either parent are placed into gametes during meiosis in the F1 generation. The F1 can be thought of as having been formed from two haploid “input” gametes.
TRUE or FALSE?
To construct a mapping cross of linked genes, it is important that the genotypes of some of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny.
To construct a mapping cross of linked genes, it is important that the genotypes of all of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny, taking into consideration that the homozygote produced only recessive gametes. Gametes and their genotypes can never be observed directly.
In a three‑point mapping experiment for the genes y‑w‑ec, the following percentages of events are observed: NCO events: 65%; SCO events between y and w: 15%; SCO events between y and ec: 17%; DCO events: 3% What is the map distance between y and ec?
20 map units
The map distance between any two genes is the sum of the percentages of all detectable recombination events between them, so 17 + 3 = 20.
For linked genes A, B, and C, the map distance A–B is 5 map units and the map distance B–C is 25 map units. If there are 10 double crossover events out of 1000 offspring, what is the interference?
0.2
The expected frequency of double crossovers would be 0.05 X 0.25 = 0.0125.
The observed double crossover frequency is 10/1000 = 0.010.
The coefficient of coincidence is 0.01/0.0125 = 0.8, so the interference is 1 – 0.8 = 0.2.
So, the interference is 1 - (0.01/0.0125) = 0.2
Suppose that 3 tomato genes from Part A did not assort independently, but instead were linked to one another on the same chromosome. Would you expect the phenotypic ratio in the offspring to change? If so, how?
All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes.
Because all three genes are linked, it is more likely that the parental allele combinations would stay together rather than be recombined through a crossover event. That is why a greater proportion of the offspring would have parental phenotypes. Nevertheless, some crossing over would likely occur, which is why a small proportion of the offspring would have recombinant phenotypes.
The recombination frequencies between genes can be used to construct a linkage map, as you have just done. The closer two genes are to each other on the same chromosome, the less likely they are to be separated by a crossover event, resulting in a lower recombination frequency.
To calculate the distance between m and p, you would add the map distances between m and d and between d and p (12 cM + 5 cM = 17 cM). Noticed that the recombination frequency between m and p is slightly less than that sum. This is because of double crossover events–the times that crossovers occur both between m and d and between d and p. In a double-crossover event, the second crossover effectively “cancels out” the first, reducing the number of recombinants between m and p that are observed, while contributing to the number of recombinants between each of the other two pairs of genes. Therefore, adding the smaller map distances to calculate larger distances avoids inaccuracies due to double-crossover events that produce non-recombinant genotypes.
The results of the Stern experiment supported the general idea that _______.
genetic recombination is a result of physical exchange between homologous chromosomes
Whenever recombinant phenotypes occurred, the cytological markers indicated that a physical exchange between the X chromosomes had also occurred.
The steps described in this tutorial are a reliable way to map loci with a testcross:
Determine the input gametes that formed the F1 dihybrid, trihybrid, or other hybrid.
Examine the testcross progeny to determine the output gametes produced by the F1.
Determine if the output gametes are recombinant or nonrecombinant for all pairs of loci.
Calculate recombination frequencies between loci by calculating the proportion of the recombinant gametes for all pairs of loci.
Use the recombination frequencies to assemble a recombination (linkage) map. Add smaller map distances to calculate larger intervals, avoiding inaccuracies due to double-crossover events that produce nonrecombinant genotypes.
Note that this example had all 3 loci linked on one chromosome, but this may not always be the case. A 3-point testcross with only two linked loci would give 2 locus pairs with a recombination frequency of 50% and one pair with a frequency less than 50%. If all 3 loci were unlinked, all three frequencies would be 50%. In all cases, however, the steps outlined in this tutorial will work to map the loci.
The discernible difference in length between the two X chromosomes of the female fruit fly was important in Stern’s experiments because _______.
it allowed cytological detection of physical exchange between the chromosomes
The differences in structure between the two chromosomes allowed Stern to track the inheritance of recombinant and nonrecombinant X chromosomes.
