Zoo/Bot 3333 Genetics Quiz 2 September 28, 2012 For answers to the quiz, click here: The M and S genes encode two protein products that form a functional heterodimer. Both protein products are needed for function, although only low threshold levels of protein are needed and the alleles at these loci are haplosufficient. An M locus mutation (Ma), eliminates the ability to pair with the wild type S gene product. A suppressor mutation, however, has arisen at the S locus (Sa), that can effectively lead to heterodimer formation with the Ma mutant, and restore function. This mutant, however, is incapable of pairing with the wild type S gene product to form a functional heterodimer; it can only interact with the Ma gene product. 1. In a cross between two MMaSSa heterozygotes, what fraction of the progeny will produce functional heterodimers? a) 14/16; b) 9/16; c) 7/16; d) 3/16; e) 1/16. 2. The F2 progeny for a dihybrid cross yields a phenotypic ratio of 9 green: 7 red. If an F1 individual was testcrossed, what would be the phenotypic ratio of the progeny? a) all green; b) all red; c) 1 green: 1 red; d) 3 green: 1 red; e) none of the above. Questions 3-5 pertain to the chart on the right. In Labrador retrievers, coat color may be black, brown (chocolate) or golden. Here are some of the many possibilities resulting from the matings of dogs of different coat colors: 3: True or false. The allele responsible for golden coat color is located at the same genetic locus as the allele for brown coat color. 4. In the cross shown in (f): a) the first parent is heterozygous for the brown allele; b) the second parent is heterozygous for the golden allele; c) the second parent is heterozygous for the black allele; d) all of the above; e) none of the above. 5. The pattern of inheritance seen in these matings indicates that the golden phenotype is an example of: a) complementary gene action; b) duplicate genes; c) dominant epistasis; d) recessive epistasis; e) dominant suppression. Questions 6-8 pertain to the following. In Drosophila, Lyra (Ly) and Stubble (Sb) are dominant mutations located on chromosome 3. A recessive mutation for bright-red eyes is discovered and also shown to be located on chromosome 3. A female who was heterozygous for all three mutations was crossed to a male homozygous for the bright-red (indicated br here) mutation, with the progeny classes produced at the right: 6. The map distance from the Stubble mutation to the Lyra mutation is: a) 3.4 map units; b) 4 map units; c) 13.4 map units; d) 14 map units; e) 18 map units. Phenotype Ly Sb br Ly + br Ly + + +++ + Sb + Ly Sb + + + br + Sb br Total Number 404 75 18 422 59 4 2 16 1000 7. True or false: For this particular data set, we see no evidence of interference between these three genes. 8. A testcross between an individual who is trihybrid for three different gene loci (A/a, B/b, C/c) and an individual homozygous recessive for all three loci yields the following progeny: Which of the following are true? a) The A and B loci are approximately equally distant from C; b) the A and B genes must be on the same chromosome and separated by about 25.3 map units; c) the C locus is located 12 map units from the A locus; d) all of the above; e) none of the above. aBC abC ABC AbC aBc abc ABc Abc 214 226 26 28 30 36 218 222 Questions 9 and 10 pertain to the following. The following are ordered tetrad pairs from a cross between strain cd and ++: Tetrad Class 1 2 3 4 5 6 7 c+ c+ cd +d c+ cd c+ c+ cd cd c+ ++ ++ +d +d ++ ++ c+ cd cd cd +d +d ++ +d +d ++ ++ 1 17 41 1 5 3 1 9. The distance between the c locus and the centromere is approximately: a) 7.3 map units; b) 14.5 map units; c) 15.9 map units; d) 20.3 map units; e) 31.8 map units. 10. The distance between these two genes is approximately: a) 9 map units; b) 23 map units; c) these genes do not appear to be linked to one another.