Molecular Evolution:Sex-specific mutation rates
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Contents |
Introduction
The fact that among birds, females are the heterogametic sex, makes it possible to test the replicative division hypothesis. This hypothesis maintains that most mutations arise during germline division. Avian sex chromosomes constitute an excellent model system for the hypothesis, in contrast with mammalian sex chromosomes, where it is not possible to separate the effects of male-biased mutation rate from those of reduction in the X-chromosome. The following table summarises the situation.
| Mammals | Birds | |
|---|---|---|
| Male-biased rate | Y > X | Z > W |
| Reduction in X or Z | Y > X | W > Z |
In this experiment, we test the replicative division hypothesis by sequencing introns from both sex chromosomes of three bird species: chicken (Gallus gallus), common quail (Coturnix coturnix) and wild turkey (Meleagris gallopavo). From the number of mutations since the sex chromosomes diverged, we obtain an estimate of the mutation rates in the respective chromosomes. We can then calculate αm, the ratio in the rates of change between males and females, by compensating for the fact that the Z-chromosome spends 1/3 of its time in the female germline, and 2/3 in the male germline. If the effects from male-biased mutation rate are stronger than those of reduction in Z, we will have αm > 1.
Methods
Lab part
PCR amplification. The first day was spent amplifying the introns the lab. Each group was instructed to amplify one of the introns 10, 11 or 25 in both of the genes CHD1Z and CHD1W. Our group amplified intron 11. This involved preparing master mixes, consisting of the following ingredients:
| Stock conc. | Conc W | 20 µl rx W | 5 rx W | |
|---|---|---|---|---|
| Buffer Gold | 10 x | 1 x | 2 µl | 10 µl |
| MgCl2 | 20 mM | 3 mM | 2.4 µl | 12 µl |
| Forw | 10 µM | 0.25 µM | 0.5 µl | 2.5 µl |
| Rev | 10 µM | 0.25 µM | 0.5 µl | 2.5 µl |
| dNTP | 20 mM | 0.2 mM | 0.2 µl | 1 µl |
| AmpliTaqGold | 5 U/µl | 0.025 U/µl | 0.1 µl | 0.5 µl |
| ddH2O | Up to 20 µl | 9.3 µl | 46.5 µl |
| Stock conc. | Conc Z | 20 µl rx Z | 5 rx Z | |
|---|---|---|---|---|
| Buffer Gold | 10 x | 1 x | 2 µl | 10 µl |
| MgCl2 | 20 mM | 2 mM | 1.6 µl | 8 µl |
| Forw | 10 µM | 0.25 µM | 0.5 µl | 2.5 µl |
| Rev | 10 µM | 0.25 µM | 0.5 µl | 2.5 µl |
| dNTP | 20 mM | 0.2 mM | 0.2 µl | 1 µl |
| AmpliTaqGold | 5 U/µl | 0.025 U/µl | 0.1 µl | 0.5 µl |
| ddH2O | Up to 20 µl | 10.1 µl | 50.5 µl |
Electrophoresis. We made a gel cast, and loaded samples from each PCR purification (mixed with loading buffer) in the wells. Then we let the electrophoresis run for 30 minutes at 150 V. We stained the gel in an ethidium bromide bath and photographed it with a UV-camera. The photo can be found in the Results section. Only the Z introns were successfully amplified.
Purification. During the electrophoresis, we also prepared for the next step by purifying the PCR products according to the QIAquick protocol.
Sequencing. We prepared one forward and one reverse sequencing reaction for each of our six samples. Three of these (the ones that indicated success in the UV photograph) were sequenced on the second day by a MegaBACE™ system.
Computer part
Editing sequences. Using ProSeq, we first went through the sequences by eye, editing out the ends where quality was bad and background noise was too prominent. We also looked for ambiguities and double peaks.
Creating contigs. After reverse-complementing the Z sequences, we assembled each W/Z sequence pair into a contig. This revealed further ambiguities in some bases.
Alignment. It turned out that alignment had been disabled in a recent version of ProSeq, so we skipped this part. But we have done it in previous courses, so the practice is not unknown to us.
Calculation. Using a maximum-likelihood approach implemented in PAML on the alignments, we estimated the number of substitutions in the introns. The results are given below. We also made a relative rate test (using MEGA) to see if there were any significant underlying mutation rate differences among the species. These results can also be found below.
Results
Gel electrophoresis
Image:Gel electrophoresis.png We made two photographs; the one to the right is the second one. (The first one shows the same results, but less clearly.) The upper part of the photo belongs to another group.
Only the three leftmost wells contained sucessfully amplified DNA. We had labeled the wells in the following order, from the left:
| Well | Species | Chromosome |
|---|---|---|
| 1 | chicken | Z |
| 2 | quail | Z |
| 3 | turkey | Z |
| 4 | null sample | |
| 5 | chicken | W |
| 6 | quail | W |
| 7 | turkey | W |
| 8 | null sample |
That is, only the PCR reactions for the CHD1Z gene were successful.
Mutation rates
These are the mutation rates that resulted from the alignments.
| total | chicken | quail | turkey | |
|---|---|---|---|---|
| 10 W | 0.131 | 0.032 | 0.058 | 0.041 |
| 10 Z | 0.194 | 0.057 | 0.070 | 0.067 |
| 11 W | 0.144 | 0.039 | 0.054 | 0.051 |
| 11 Z | 0.212 | 0.064 | 0.091 | 0.057 |
| 25 W | 0.091 | 0.009 | 0.050 | 0.032 |
| 25 Z | 0.208 | 0.043 | 0.087 | 0.078 |
| All W | 0.120 | 0.026 | 0.054 | 0.040 |
| All Z | 0.203 | 0.054 | 0.083 | 0.076 |
From this, the male to female mutation rate ratios αm can be calculated by
| Image:Male to female mutation rate ratio formula.png |
which yields the following rates:
| total | chicken | quail | turkey | |
|---|---|---|---|---|
| 10 | 1.71 | 2.19 | 1.29 | 1.93 |
| 11 | 1.71 | 1.96 | 2.03 | 1.17 |
| 25 | 2.92 | 6.60 | 2.12 | 3.13 |
| All | 2.03 | 2.58 | 1.80 | 1.98 |
Here are the same results shown as a histogram:
| Image:Mutation rate chart .png |
Relative rate test
According to the results of the relative rate tests on the W chromosomal data, we had the following relative rates:
(where "<" should be read as "has a slower mutation rate than".) The tests on the Z data were inconclusive (not all rate differences were statistically significant), but yielded that
which is at least consistent with the W data.
Discussion
Mutation rate variation. The male/female ratios in the results are significantly and consistently above 1, which seems to corroborate the replicative division hypothesis. No particular species has a higher ratio across all the introns. The surprisingly high value for intron 25/chicken may be due to poor alignment or other measurment error; even so, the ratios for intron 25 are clearly higher than those for 10 and 11. This is something that we are unable to account for except by random rates in mutation along the chromosome.
Generation time effects. No mutation rate variation due to differences in generation time can be observed. Then again, this paper suggests that galliform species (which includes chicken, quail and turkey) all have an average generation time of 2 years. To the extent that this can be interpreted as the three species having fairly similar generation times, maybe we should not excpect to see any such effects.
What kinds of sequences are suitable for studying mutation rate variation? We want sequences where random mutations are not subject to selection. Introns are a good example of this, as long as we make sure that they really have no selective significance. The Wikipedia article on noncoding DNA suggests that they sometimes do.
