The genes for sepia eye color, short bristles, and dark body coloration are on the same chromosome of Drosophila melanogaster. Each gene has two alleles: wild type, which is dominant, or mutant, which is recessive.
se+ is dominant and causes red eyes; se is recessive and causes sepia eyes
sb+ is dominant and causes long bristles; sb is recessive and causes short bristles
b+ is dominant and causes gray body coloration; b is recessive and causes dark body coloration
The sepia gene and short bristle gene are separated by 32 map units; the short bristlegene and the dark body gene are separated by 13 map units. The sepia gene and dark bodygene are separated by 45 map units.
If you crossed a fly, which was homozygous for the wild type allele of each gene with a homozygous recessive fly for each gene, the F1 generation would entirely consist of flies that had the wild type traits, but were heterozygous for each gene.
If you were to cross these heterozygous F1flies to recessive homozygous flies, how many flies would you predict to have the double cross-over phenotype? (You will show all of your work and phenotypic classes in the next problem. Just show the NUMBER of double cross-over flies predicted by the map unit data here. Round to a whole number)
If you were to cross these heterozygous F1flies to recessive homozygous flies, how many flies would you predict to observe in each of the eight predicted phenotypic categories with respect to these three genes (given the map units above) if there were 1,000 total offspring? Please show work and show all phenotypic classes!
The genes for sepia eye color, short bristles, and dark body coloration are on the same chromosome of Drosophila melanogaster. Each gene has two alleles: wild type, which is dominant, or mutant, which is recessive.
se+ is dominant and causes red eyes; se is recessive and causes sepia eyes
sb+ is dominant and causes long bristles; sb is recessive and causes short bristles
b+ is dominant and causes gray body coloration; b is recessive and causes dark body coloration
The sepia gene and short bristle gene are separated by 32 map units; the short bristlegene and the dark body gene are separated by 13 map units. The sepia gene and dark bodygene are separated by 45 map units.
If you crossed a fly, which was homozygous for the wild type allele of each gene with a homozygous recessive fly for each gene, the F1 generation would entirely consist of flies that had the wild type traits, but were heterozygous for each gene.
If you were to cross these heterozygous F1flies to recessive homozygous flies, how many flies would you predict to have the double cross-over phenotype? (You will show all of your work and phenotypic classes in the next problem. Just show the NUMBER of double cross-over flies predicted by the map unit data here. Round to a whole number)
If you were to cross these heterozygous F1flies to recessive homozygous flies, how many flies would you predict to observe in each of the eight predicted phenotypic categories with respect to these three genes (given the map units above) if there were 1,000 total offspring? Please show work and show all phenotypic classes!
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Related questions
In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u.. long. The X-linked gene for body color-with two alleles, y +for gray body and y for yellow body-resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w +for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f + for normal bristles and f for forked bristles, is located at map position 56.7. Each gene resides on the Xchromosome, and at each locus the wild-type allele is dominant over the mutant allele.
Predict the frequency of gray with white eyes and forked bristles progeny type produced by this mating.
Enter your answer as a percentage to three decimal places (example 0.235 or 0.230).
Frequency=_________
2.
| Phenotype | Number |
| Pale | 648 |
| Pale, oval | 64 |
| Pale, short | 10 |
| Pale, oval, short | 102 |
| Oval | 6 |
| Oval, short | 618 |
| Short | 84 |
| Wild type | 98 |
| 1630 |
A wild-type trihybrid soybean plant is crossed to a pure-breeding soybean plant with the recessive phenotypes pale leaf l, oval seed (r), and short height (t). The results of the three-point test cross are shown in the table. Traits not listed are wild type.
Calculate the interference value for these data.
Enter your answer to three decimal places (example 0.201).
