Genetics 371B Practice problems--Autumn 2000 week 10
Lectures 30-34
| 1. | The second thoracic segment in Drosophila is supposed to produce wings, while the third thoracic segment is supposed to produce halteres (flight balancers). Suppose that Gene D is needed for proper specification of the second thoracic segment, and Gene E is needed for proper specification of the third thoracic segment. Based on what you know about segmental identity is set, state whether you expect each of the following phenotypes to result from a loss of function mutation in Gene D or Gene E, or a gain of function mutation in either of these two genes: | |
| (a) | Both second and third thoracic segments produce wings | |
| (b) | Both second and third thoracic segments produce halteres | |
| 2. | Exam question from 1998
You identify two transcription factors, Muscle1 (M1) and Muscle2 (M2), that are transcribed only in muscle cells. If you artificially express M1 or M2 in a cultured skin cell, it turns into a muscle cell. |
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| (a) | m1 and m2 mutations are both recessive embryonic lethal as a result of not having any muscle cells. In m1 mutant embryos the cells that should have become muscle express M2, but in m2 mutant embryos M1 is never expressed. Draw a pathway for how M1 and M2 control muscle differentiation. Explain briefly (3 sentences) why your pathway is the only one that makes sense. | |
| (b) | If you artificially induce (using some molecular trick) a brief pulse of exogenous M2 expression in a skin cell, it becomes a muscle cell forever, whereas if you artificially induce a brief pulse of exogenous M1 expression in a skin cell it briefly takes on a muscle-like appearance but soon becomes a skin cell again. Without invoking any other gene or gene product, give a hypothesis to explain this observation, and comment briefly (one sentence) on the developmental significance of this phenomenon. | |
| 3. | Keeping in mind that oligonucleotide microarrays (see p.59 of the lecture notes) can be designed such that a single base mismatch can prevent hybridization of probe, speculate on how microarrays may be used to screen for disease alleles in newborns. Specifically, what DNA sequences would you want to have on the arrays, and what would you hybridize tohe arrays? | |
| 4. | Gene therapy often involves introduction of a wild type copy of a gene into cells from patients with mutations in the gene. For recessive disease phenotypes, in principle, it should be sufficient to integrate the wild type allele anywhere in the target cell's genome. In practice, it is found that such integration at non-native locations (or "ectopic" sites) often results in either overexpression or silencing of the introduced gene. | |
| (a) | Suggest an explanation for this observation. | |
| (b) | Why might it be undesirable to allow overexpression of the wild type allele? | |
| (c) | What strategy would you try, to get around this problem? | |
| 5. | For each of the following pairs of populations, state whether you think one of the members of the pair will show higher heritability (in the broad sense) for trait T, and if so, which one and why: | |
| (a) | a population that is mostly homozygous for the genes controlling trait T, or one that is mostly heterozygous | |
| (b) | a population that is in a mostly uniform environment with respect to factors that affect trait T, or one that is in a more heterogeneous environment. | |
| 6. | Comparison of monozygotic twins raised together or raised apart is often touted as a way of estimating heritability of behavioral and personality traits. | |
| (a) | Suppose that monozygotic twins raised apart show some personality trait P about as frequently as do twins raised together. Does this observation prove a genetic basis for that personality trait? Can you think of (and suggest!) some non-genetic factors that could give the observed results? What would you like to see in such studies to control for the effect of non-genetic factors? | |
| (b) | How about the converse--if twins raised apart show a trait less often than do twins raised together, does that eliminate the possibility of a genetic basis to the trait? Explain. | |
| 7. | Assume that height in a plant is controlled by two gene pairs and that each additive allele contributes 10 cm to a base height of 10 cm (i.e., aabb = 10 cm). | |
| (a) | What is the height of an AABB plant? | |
| (b) | Predict the phenotypic ratios of F1 and F2 plants in a cross between aabb and AABB. | |
| (c) | List all the genotypes that give rise to plants that are 20 and 40 cm in height. | |
| 8. | In a cross where three gene pairs determine weight in squash, what proportion of individuals from the cross AaBbCC x AABbcc will contain only 2 additive alleles? What genotype(s) fall into this category? |
| 9. | An inbred strain of plants has a mean height of 24 cm. A second strain of the same species from a different country also has a mean height of 24 cm. The F1 plants from a cross between these two strains are also 24 cm high. However, the F2 generation shows a wide range of heights; the majority are like the P1 and F1 plants, but approximately 4 of 1000 are only 12 cm high, and 4 of 1000 are 36 cm high. | |
| (a) | What mode of inheritance is occurring here? | |
| (b) | How many gene pairs are involved? | |
| (c) | How much does each gene contribute to the plant height? | |
| (d) | Indicate one possible set of genotypes of the P1 and F1 plants that could explain their heights. | |
| (e) | Indicate one possible set of genotypes to account for F2 plants that are 18 cm or 33 cm high. | |