Answer key, practice problems 2000--week 10

1. Remember that the default seems to be to produce more anterior-like segments; additional genes have to be expressed to set posterior fates. Keeping that in mind --
(a) Failure to produce a posterior segment implies failure of expression of a gene specifying that segment. Therefore, this phenotype is consistent with a loss of function mutation in Gene E (the gene required for the third segment).
(b) Here, an anterior segment has taken on a posterior fate -- this phenotype is consistent with a gain of function mutation in Gene E (inappropriate expression of that gene has forced the 2nd thoracic segment to take on a 3rd segment fate).

2. (a)

Both gene products are needed for muscle development, suggesting a positive interaction (one activating the other). M1 expression is dependent on M2 but not vice versa, so M2 must act upstream of M1 (otherwise we sould see no M2 transcription in the absence of M1). An alternative possibility, that M1 represses M2, which represses muscle formation, can also be ruled out because besides being a negative interaction, M1 transcription should then be independent of M2 transcription, which is not the case.
(b)

This results suggests that M2, besides activating M1 expression, also activates its own expression in a feedback loop. Therefore, once M2 has been activated, it keeps itself (and therefore M1) active forever, permanently changing the skin cells to muscle cells. In contrast, M1 does not activate itself or M2, so a pulse of M1 mRNA gives a muscle phenotype briefly, but this expression is only transient; in the absence of sustained M1 transcription, the cell reverts to the skin cell phenotype.

These observations indicate that maintenance of muscle fate requires continuous expression of the M2/M1 pathway.

3. Currently, genetic diseases are screened one gene at a time. With oligonucleotide microarrays, it becomes possible to screen for a variety of different disease alleles simultaneously. First, we have to know the sequence of the wild type and of the mutant allele, specifically, the location of the mutation. The arrays can then be set up in two different ways. One way is to have the wild type alleles on the microarray (one allele per spot on the microarray). The whole gene would NOT be on the array; only a small snippet (25 bases is what is currently used) is made, such that the location of the mutation is in the middle of the piece. (This strategy assumes that we are screening for small mutations, although large deletions would be detected also.) DNA from the infant to be screened is labeled and hybridized to the array. Wherever we see hybridization, we can conclude that a wildtype allele was present in the infant; failure to see hybridization would be in indication of a non-wildtype allele in the infant. The alternative approach is to have the mutant allele sequences on the array; then, hybridization of DNA from the infant would indicate that the infant had the mutant allele. Both approaches are subject to the limitation that there may be polymorphisms in the "wild type" population.

One can also think of including polymorphic sites that are closely linked to disease genes, and screening for those, but that increases the danger of false negatives -- a polymorphic allele that is closely linked to a disease allele in one family may have recombined away from the disease allele in another family.

4. (a) The gene may have landed in a region that places it under control of regulatory elements of some other gene. This scenario is similar to the situation we saw earlier with Burkitt lymphoma, where a translocation of the c-myc gene is believed to place it under control of the immunoglobulin gene cluster. The idea is the same -- when a gene is placed in a non-native location, it may end up near regulatory elements that alter its expression compared to the native state.
(b) Overexpression of even the wildtype allele can be bad (although not necessarily so) -- it would in effect be a gain of function allele, which, as we have seen, can have deleterious consequences.
(c) An obvious strategy is to do an exact replacement of the mutant allele with a wiltype allele, so all of the chromosomal "context" is kept intact.

5. (a) Remember that heritability is a measure of how much of the variation can be ascribed to genetic influence. Thus, the more heterozygous population is genetically more heterogeneous, and will therefore show higher heritability than the more homozygous population. (In the homozygous population, there is relatively little genetic variation, so we have to ascribe a larger fraction of the phenotypic variation to non-genetic factors.)
(b) The population in the more uniform environment will show higher heritability.

6. (a) Not necessarily--it would depend on the trait, and how frequently the trait is observed in genetically unrelated (or less related) individuals. Environmental (including cultural) factors may play a role too--e.g., family customs and values, socioeconomic conditions, etc. One way to control for such factors is to compare monozygotic twins who have been raised in very different circumstances.
(b) Again, not necessarily--here, the environment could act against concordance (i..e, the appearance of the same trait in the twins) even if there is a genetic predisposition toward concordance. One control would be to ask if there is significantly higher concordance between twins than between unrelated individuals raised in the same pairs of conditions.

7. (a) 50 cm (10 cm per additive allele x 4 additives, added to the base height of 10 cm)
(b) The F1 plants will be 30 cm tall (genotype = AaBb, i.e., 2 additive alleles = 20 cm added to the base height of 10 cm).

F2 will be 10, 20, 30, 40, and 50 cm plants in 1:4:6:4:1 ratio; this answer can be calculated from the binomial expansion --

(x + y)2n = 1

where x = non-additive allele, y = additive, and n = # of genes = 2. (I am using 'x' and 'y' here to avoid confusion with the A/a and B/b gene names.)

Expanding the equation, we get:

x4 + 4x3y + 6x2y2 + 4xy3 + y4 = 1

Keeping in mind that the exponents (the powers of x and y) reflect the number of those alleles in that genotype (e.g., x4 --> 4 non-additive alleles, x2y2 --> 2 non-additive, 2 additive alleles, etc.) and the coefficients reflect the proportions, we get the distribution of plant heights described above.
(c)

20 cm plants have one additive allele -- genotype Aabb or aaBb.

40 cm plants have three additive alleles -- genotype AABb or AaBB.

8.
gametes ABc (2) Abc (1)
ABC (3) AABBCc (5) AABbCc (4)
AbC (2) AABbCc (4) AAbbCc (3)
aBC (2) AaBBCc (4) AaBbCc (3)
abC (1) AaBbCc (3) AabbCc (2)
The cross is AaBbCC x AABbcc. As seen from the diagram, 1/8 of the progeny will have 2 additive alleles; the genotype of this class will be AabbCc. (The numbers of additive alleles in each case is shown in parentheses.)

9. (a) Quantitative inheritance.
(b) The frequency of either extreme phenotype gives us n, the number of gene pairs--

Frequency of one extreme phenotype = (1/4)n = 1/250

# of gene pairs = log(250)/log(4) = 4.
(c) The maximum contribution of additive alleles = 36 - 12 = 24 cm. Since 8 additive alleles (4 genes) contribute 24 cm, each additive allele contributes 3 cm.
(d) Each parent has 4 additive alleles; since the F1 also have 4 additive alleles, the parents must be each be homozygous; the additive alleles of one parent are not present in the other. For example, the genotypes could be AABBccdd x aabbCCDD (or other genotypes following that pattern).
(e) An 18 cm plant has 2 additive alleles; any genotype such as AAbbccdd or aaBBccdd would work. A 33 cm plant has 7 additive alleles; any genotype such as AABBCCDd or AaBBCCDD would work.