Genetic Interactions Practice Problems (page 3)
Review the following concepts if needed:
- Two-Factor Interactions for Genetics
- Epistatic Interactions for Genetics
- Pedigree Analysis Help
- Nonepistatic Interactions and Pleiotropism for Genetics
Genetic Interactions Practice Problems
Coat colors of dogs depend upon the action of at least two genes. At one locus a dominant epistatic inhibitor of coat color pigment (I-) prevents the expression of color alleles at another independently assorting locus, producing white coat color. When the recessive condition exists at the inhibitor locus (ii), the alleles of the hypostatic locus may be expressed, iiB- producing black and iibb producing brown. When dihybrid white dogs are mated together, determine (a) the phenotypic proportions expected in the progeny, (b) the chance of choosing, from among the white progeny, a genotype that is homozygous at both loci.
- The genotypic proportions among the white progeny are as follows:
The only homozygous genotypes at both loci in the above list are 1/12 IIBB and 1/2 IIbb = 2/12 or 1/6 of all the white progeny. Thus there is 1 chance in 6 of choosing a homozygous genotype from among the white progeny.
Matings between black rats of identical genotype produced offspring as follows: 14 cream-colored, 47 black, and 19 albino. (a)What epistatic ratio is approximated by these offspring? (b)What are the genotypes of the parents and the offspring (use your own symbols)?
- The total number of offspring is 80 (14 + 47 + 19) and 80/16 = 5. So 1/16 of 80 = 5. In order to figure out what the ratio is, use this factor. So, 5 × 3 = 15, which is close to 14; 5 × 4 = 20, which is close to 19; and 5 ×9 = 45, which is close to 47. Thus, the ratio is 9 : 3 : 4 (the numbers are not given in this order). This is an example of recessive epistasis.
This translates to a 9 black : 3 cream : 4 albino ratio; cc, the recessive condition, is epistatic to alleles at locus B.
Red color in wheat kernels is produced by the genotype R-B-, white by the double-recessive genotype (rrbb). The genotypes R-bb and rrB- produce brown kernels. Ahomozygous red variety is crossed to awhite variety. (a) What phenotypic results are expected in the F1 and F2? (b) If the brown F2 is artificially crossed at random (wheat is normally self-fertilized), what phenotypic and genotypic proportions are expected in the offspring?
- The proportion of genotypes represented among the brown F2 must first be determined.
Next, the relative frequencies of the various matings may be calculated in a Punnett square.
A plant of the genus Capsella, commonly called "shepherd's purse," produces a seed capsule, the shape of which is controlled by two independently assorting genes, represented by symbols A and B. When dihybrid plants were interpollinated, 6% of the progeny were found to possess ovoid-shaped seed capsules. The other 94% of the progeny had triangular-shaped seed capsules. (a)What two-factor epistatic ratio is approximated by the progeny? (b) What is the genotype of the ovoid-shaped seed capsules?
- Ninety-four percent of 16 is ~ 15 (0:94 × 16 = 15:04; or 15/16 × 100 = 93:75) and 6% of 16 is ~ 1, thus, the ratio represented here is 15 triangular : 1 ovoid
- The genotype is the double recessive, aabb, which occurs 1/16 of the time. The dominant condition A- or B- is epistatic to either recessive condition, so A-B- (9/16), A-bb (3/16), aaB- (3/16) all exhibit the dominant phenotype (9 + 3 + 3 = 15).
Two white-flowered strains of the sweet pea (Lathyrus odoratus) were crossed, producing an F1 with only purple flowers. Random crossing among the F1 produced 96 progeny plants, 53 exhibiting purple flowers, and 43 with white flowers. (a) What phenotypic ratio is approximated by the F2? (b) What type of interaction is involved? (c) What were the probable genotypes of the parental strains?
- To determine the phenotypic ratio in terms of familiar sixteenths, the following proportion for white flowers may be made: 43/96 = x/16, from which x = 7.2. That is, 7.2 white : 8.8 purple, or approximately a 7 : 9 ratio. We might just as well have arrived at the same conclusion by establishing the proportion for purple flowers: 53/96 = x/16, from which x = 8.8 purple.
- A 7 : 9 ratio is characteristic of duplicate recessive genes where the recessive genotype at either or both of the loci produces the same phenotype.
- If aa or bb or both could produce white flowers, then only the genotype A-B- could produce purple. For two white parental strains (pure lines) to be able to produce an all-purple F1, they must be homozygous for different dominant-recessive combinations. Thus,
A dominant gene S in Drosophila produces a peculiar eye condition called "star." Its recessive allele S+ produces the normal eye of wild type. The expression of S can be suppressed by the dominant allele of another locus, Su-S. The recessive allele of this locus, Su-S+,has no effect on S+. (a) What type of interaction is operative? (b) When a normal-eyed male of genotype Su-S / Su-S, S/S is crossed to a homozygous wild-type female of genotype Su-S+ / Su-S+, S+ / S+, what phenotypic ratio is expected in the F1 and F2?
- Dominant and recessive interaction; both dominant and recessive alleles interact to result in an eye phenotype
All the F1 offspring are wild type, due to the presence of the dominant Su-S allele. F2: 3/16 S–/-, Su-S+ / Su-S+ star eye: 13/16 wild type.
The following pedigree involving two-factor epistasis shows the transmission of swine coat colors through three generations:
Assume the offspring of II5 × II6 shown in this pedigree occur in the ratio expected from the genotypes represented by their parents. How are these colors most likely inherited?
Notice first that three phenotypes are expressed in this pedigree. This rules out epistatic combinations producing only two phenotypes such as those expressed in dominant and recessive interaction (13 : 3), duplicate dominant genes (15 : 1), and duplicate recessive genes (9 : 7). Epistatic gene interactions producing three phenotypes are those expressed in dominant epistasis (12 : 3 : 1), recessive epistasis (9 : 3 : 4), and dominant genes with cumulative action (9 : 6 : 1). Let us proceed to solve this problem by making an assumption and then applying it to the pedigree to see if the phenotypes shown there can be explained by our hypothesis.
Case 1. Assume dominant epistasis is operative. The genotypes responsible for the three phenotypes may be represented as follows: A-B- and A-bb = first phenotype, aaB- = second phenotype, and aabb = third phenotype. We must now determine which of the phenotypes represented in this pedigree corresponds to each of the genotypic classes. Obviously the only pure-line phenotype is the third one. Offspring of the mating aabb × aabb would all be phenotypically identical to the parents. The mating I4 × I5 appears to qualify in this respect and we shall tentatively assume that white coat color is represented by the genotype aabb. Certain matings between individuals with the dominant epistatic gene A could produce three phenotypically different types of offspring (e.g., AaBb × AaBb). Such a mating is observed between II2 and II3. Therefore, we might assume red color to be represented by the genotype A-. Sandy color must then be represented by genotype aaB-. Matings between sandy individuals could produce only sandy (aaB-) or white (aabb) progeny. However, sandy parents I1 × I2 produce white and red progeny (II1, II2). Therefore the assumption of dominant interaction must be wrong.
Case 2. Assume recessive epistasis to be operative. The genotypes responsible for the three phenotypes in this case may be represented as follows: A-B- as first phenotype, A-bb as second phenotype, and aaB- and aabb as third phenotype. As pointed out in case 1, matings between individuals of genotype AaBb are the only kind among identical phenotypes that are capable of producing all three phenotypes in the progeny. Thus A-B- should represent red (e.g., II2 × II3). The aa genotypes breed true, producing only white individuals (I4 × I5). Sandy is produced by genotype A-bb. Sandy × sandy (I1 × I2) could not produce the red offspring (II2). Therefore the assumption of recessive interaction must be wrong.
Case 3. Assume that duplicate genes with cumulative action are interacting. The genotypes responsible for the three phenotypes in this case may be represented as follows: A-B- as first phenotype, A-bb and aaB- as second phenotype, and aabb as third phenotype. As explained in the previous two cases, A-B- must be red and aabb must be white. If we assume that any dominant genotype at either the A locus or B locus contributes one unit of pigment to the phenotype, then either the genotype aaB- or A-bb could be sandy; we further assume that the presence of both dominant genes (A-B-) would contribute two units of pigment to produce a red phenotype. Thus, the mating II5 (AaBb) red × II6 (aaBb) sandy would be expected to produce offspring phenotypes in the following proportions:
The same phenotypic ratio would be expected if II6 were Aabb. These expectations correspond to the ones given in the pedigree (III4–III11) and therefore the hypothesis of duplicate genes with cumulative action is consistent with the data.
At least three loci are known to govern coat colors in mice. The genotype C- will allow pigment to be produced at the other two loci. The recessive genotype cc does not allow pigment production, resulting in "albino." The "agouti" pattern depends upon the genotype A-, and nonagouti upon the recessive aa. The color of the pigment may be black (B-) or chocolate (bb). Five coat colors may be produced by the action of alleles at these three loci:
- What phenotypic frequencies are expected in the F2 from crosses of pure black with albinos of type AAbbcc?
- A cinnamon male is mated to a group of albino females of identical genotype and among their progeny were observed 43 wild type, 40 cinnamon, 39 black, 41 chocolate, and 168 albino. What are the most probable genotypes of the parents?
In part (b), the cinnamon progeny, A-bbC-, indicate b in the female parents. The black progeny, aaB-C-, indicate a in both parents, and B in the female parents. The chocolate progeny, aabbC-, indicate a in both parents, and b in the females. The albinos indicate c in themale. The genotype of themale is now known to be AabbCc. But the genotype of the albino females is known only to be a-Bbcc. They could be either AaBbcc or aaBbcc
Lewis-a blood group substance appears on the human red blood cell when the dominant gene Le is present, but is absent if the dominant gene of the "secretor" locus Se is present. Suppose that from a number of families where both parents are Lewis-a negative of genotype LeleSese, we find that most of them have 3 Lewis-a positive : 13 Lewis-a negative children. In a few other families, suppose we find 2 Lewis-a negative : 1 Lewis-a positive. Furthermore, in families where both parents are secretors of genotype Sese, we find most of them exhibit a ratio of 3 secretor : 1 nonsecretor, but a few of them show 9 secretor : 7 nonsecretor. Propose a hypothesis to account for these results.
If only two loci are interacting, the dominant Se gene can suppress the expression of Le, resulting in Lewis-a negative blood type. When both parents are dihybrid, we expect a 13 : 3 ratio in the progeny, characteristic of dominant and recessive interaction.
The 9 : 7 ratio found in some families for the secretor trait indicates that two factors are again interacting. This is the ratio produced by duplicate recessive interaction; i.e., whenever the recessive alleles at either of two loci are present, a nonsecretor phenotype results. Let us symbolize the alleles of the second locus by X and x.
If we assume the x gene to be relatively rare, then most families will have only the dominant gene X, but in a few families both parents will be heterozygous Xx. Let us assume that this is the case in those families that produce 2 Lewis-a negative : 1 Lewis-a positive.>
If Se suppresses Le, but xx suppresses Se, then only the genotypes marked with an asterisk (*) will be Lewis-a positive, giving a ratio of 21 Lewis-a positive : 43 Lewis-a negative. This is very close to a 1 : 2 ratio and indeed would appear to be such with limited data.
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