Questions & explanations
1. Given the following data from a three-point test cross: a b c 270, a b + 10, a + c 60, a + + 15, + b c 20, + b + 55, + + c 5, + + + 265. Determine the gene order.
First, identify the parental types (most frequent): a b c and + + + (270 and 265). The double crossover types (least frequent): a + + and + b c (15 and 20? Actually, check: least frequent are a + + (15) and + b c (20)? But also a b + (10) and + + c (5) are even fewer? Wait, let's list: a b c 270, a b + 10, a + c 60, a + + 15, + b c 20, + b + 55, + + c 5, + + + 265. The smallest are + + c (5) and a b + (10). These are the double crossover classes. Compare them to parental: parental a b c and + + +. The double crossover a b + differs from a b c at the c locus (c vs +), and from + + + at a and b? Actually, to find order, compare double crossover with parental: For double crossover a b +, if order is a-b-c, then a and b are same as parental a b c, and c is different. That suggests c is the middle gene? Let's do systematically: The double crossover classes are a b + and + + c. Compare to parental a b c: a b + differs at c. Compare to parental + + +: + + c differs at c. So the gene that differs in both double crossovers is c. Therefore, c is the middle gene. So order is a-c-b or b-c-a. Che
2. In a test cross, you get 45% parental and 5% recombinant for genes A and B, but 30% parental and 20% recombinant for genes A and C. Which pair of genes is closer?
The pair with the lower recombination frequency is closer. For A and B, recombination frequency is 5% (since 5% recombinants out of 50% total? Actually, in a test cross, parental and recombinant percentages sum to 100%? Wait, careful: In a test cross, if you have 45% parental and 5% recombinant for A and B, that sums to 50%? That can't be right. Typically, you count total offspring. Let's correct: Suppose total offspring = 100, then for A-B: 45 parental + 5 recombinant = 50? That implies 50% are something else? Actually, in a two-point test cross, there are two parental types and two recombinant types. So if one parental type is 45%, the other parental is also 45%? No, usually they are equal. Let's assume the numbers are percentages of total offspring. For A-B: 45% parental (both types combined) and 5% recombinant (both types combined) gives 50% total? That's not possible. I think the intended meaning: recombination frequency = recombinants/total. So for A-B, recombinants = 5% of total, so RF=5%. For A-C, recombinants=20%, RF=20%. So A and B are closer because 5% < 20%. So answer: A
3. How does realized heritability differ from narrow-sense heritability estimated from pedigrees?
Realized heritability is directly observed from selection experiments, while narrow-sense heritability is estimated from resemblance between relatives. Realized heritability uses the actual response to selection, so it reflects the additive genetic variance that responded. Narrow-sense heritability includes all additive variance, but may include non-additive parts if not properly estimated. Both are ratios of additive genetic variance to phenotypic variance. Realized heritability can be more accurate if the experiment is well-controlled. However, it only applies to the specific selection method and population used. Pedigree-based estimates can be generalized to the whole population.
4. What is selection differential?
Selection differential is the difference between the mean of the selected parents and the mean of the whole population before selection. For example, if the average height in a plant population is 100 cm and you select plants with average height 120 cm to be parents, the selection differential is 20 cm. It measures how much better the chosen parents are than the average. Selection intensity is the selection differential divided by the phenotypic standard deviation. It standardizes the selection differential, allowing comparison across traits with different units. If the standard deviation is 10 cm, then selection intensity = 20/10 = 2. Higher intensity means stronger selection.
5. What does Tajima's D measure in a DNA sequence sample?
Tajima's D compares two ways to estimate the mutation rate: one from the number of different DNA letters (segregating sites) and another from the average number of differences between pairs of sequences (nucleotide diversity). If the two estimates are equal, D is zero, meaning the population is evolving as expected under neutral theory. A negative D means there are more rare variants than expected, which can happen after a population expansion or a selective sweep. A positive D means there are fewer rare variants, which can happen after a population bottleneck or balancing selection. So Tajima's D helps detect if natural selection or population size changes have occurred.
6. What is the difference between one-sample and two-sample Mendelian randomization?
In one-sample MR, you have data on the genetic variant, the risk factor, and the outcome all from the same group of people. You can directly test the association between the variant and the risk factor, and between the variant and the outcome. In two-sample MR, you use summary statistics from two separate studies: one that gives the effect of the variant on the risk factor (e.g., from a GWAS of BMI), and another that gives the effect of the variant on the outcome (e.g., from a GWAS of heart disease). Two-sample MR is more common because large GWAS summary data are publicly available, but it requires that the two samples come from similar populations.
7. Compare Fay and Wu's H with Tajima's D in terms of the type of selection they detect.
Tajima's D detects deviations from neutrality in the allele frequency spectrum, such as an excess of rare variants (negative D) from population expansion or selective sweep, or an excess of intermediate variants (positive D) from balancing selection. Fay and Wu's H specifically detects an excess of high-frequency derived alleles, which is a strong signal of a recent selective sweep. While both can be negative under a sweep, H is more sensitive to sweeps because it focuses on the high-frequency end of the spectrum. Tajima's D might also be negative due to population growth, but H is less affected by growth and more specific to selection.
8. Discuss the ethical considerations of using gene editing to modify quantitative traits in livestock, compared to conventional breeding.
Conventional breeding uses natural reproduction and selection, which is generally accepted. Gene editing raises concerns about animal welfare if edits cause unintended harm. For quantitative traits, editing multiple genes might have unpredictable side effects. There is also worry about reducing genetic diversity, making populations vulnerable to new diseases. Additionally, gene editing in livestock could lead to patenting of animals, affecting farmers' autonomy. Public acceptance varies; some see it as unnatural. Regulatory frameworks differ by country, with some banning gene-edited animals for food. These issues require careful debate.
9. Compare gene editing with traditional genomic selection for improving growth rate in beef cattle. Which is more sustainable?
Genomic selection uses natural variation and accumulates small beneficial alleles over generations through mating. It is sustainable because it maintains genetic diversity and avoids unintended effects. Gene editing can introduce specific changes quickly, but it may disrupt other traits due to pleiotropy. For growth rate, editing a single gene might increase muscle but reduce fertility. Genomic selection improves all traits simultaneously using a selection index. Also, gene editing requires regulatory approval and public acceptance, which can be barriers. Thus, genomic selection is currently more sustainable for long-term improvement.
10. What is Mendelian randomization (MR)?
Mendelian randomization is a method that uses genetic variants as tools (instrumental variables) to figure out whether a risk factor actually causes a disease. It works like a natural randomized trial: because genes are assigned at conception and not changed by lifestyle, they can help separate cause from effect. For example, if a gene variant makes people have higher LDL cholesterol, and that variant is also linked to heart disease, it suggests LDL cholesterol causes heart disease. MR avoids some problems of observational studies, like reverse causation (disease causing the risk factor) and confounding (other factors affecting both).
11. What are the consequences of field trial destruction for scientists and companies?
Field trial destruction can cause significant financial losses for scientists and companies. Years of research and investment can be lost in a single night. It also delays the development of potentially beneficial crops, such as those resistant to drought or disease. Scientists may become more secretive or move their trials to other countries with stronger security, which can reduce public oversight. Additionally, destroyed trials mean that safety data cannot be collected, which may actually increase risks if the crop is later released without proper testing. The destruction also undermines trust between researchers and the public.
12. A company develops a GM potato that produces less acrylamide when fried. How would the approval process differ in the US vs the EU?
In the US, the company would likely go through a voluntary consultation with the FDA to show the potato is safe and not significantly different from conventional potatoes. If the FDA agrees, the potato can be marketed without special labeling. In the EU, the company must submit a full application to the European Food Safety Authority (EFSA) for a scientific risk assessment. The application includes molecular characterization, toxicity tests, and environmental impact studies. After EFSA's opinion, the European Commission and member states vote on authorization, which can take years. The potato would also need to be labeled as GM.