1. A purple flowering, smooth seed dihybrid plant (genotype PpSs) is test crossed with a white flowering, wrinkled seed plant (genotype ppss). These produce progeny in the following numbers of four phenotypes: 14:86:84:16 (purple flower + smooth seed coat: purple flower + wrinkled seed coat: white flower + smooth seed coat: white flower + wrinkled seed coat).
a) What is the expected ratio of progeny phenotypes assuming independent assortment of alleles?
b) Explain how the actual ratio of progeny (above) shows that the two genes are linked.
c)How many map units separate the colour and seed coat genes? Show your calculations.
1. A purple flowering, smooth seed dihybrid plant (genotype PpSs) is test crossed with a white flowering, wrinkled seed plant (genotype ppss). These produce progeny in the following numbers of four phenotypes: 14:86:84:16 (purple flower + smooth seed coat: purple flower + wrinkled seed coat: white flower + smooth seed coat: white flower + wrinkled seed coat).
a) What is the expected ratio of progeny phenotypes assuming independent assortment of alleles?
b) Explain how the actual ratio of progeny (above) shows that the two genes are linked.
c)How many map units separate the colour and seed coat genes? Show your calculations.
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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).
Which of these statements is incorrect?
Syntenic genes are located on the same chromosome. |
Independent assortment results in recombinant chromosomes. |
You can reliably predict the relative genetic distance fromgenesâ physical distance on a chromosome. |
Linked genes are always syntenic. |
What is the relative genetic distance between two linked genesif the recombination frequency is 0.49?
0.49 cM |
4.9 cM |
49 cM |
490 cM |
What statement best explains the distortion in Mendelian ratiosobserved by Bateson & Punnett in 1905? (Reminder: they found anoverrepresentation of F2 offspring showing both dominant orrecessive phenotypes, and an underrepresentation of offspringdisplaying one dominant and one recessive phenotype)
| Human error: they should have been more careful about theirexperimental setup. |
| Gene linkage: Genes for flower color and pollen shape arephysically close on the same chromosome, leading to a breakdown inthe independent assortment of the alleles for these traits. |
| Chromosome crossover: Homologous recombination of twochromatids during meiosis caused the alleles to shuffle, resultingin a breakdown of the independent assortment of the alleles forthose genes. |
Random variation: No two situations are alike. In finitepopulations, you are going to get some variation across a mean. |
When determining the relative genetic distance between twogenes, why is dihybrid back-cross preferable over traditionaldihybrid cross?
| 9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1ratio. |
| Genotypes of the offspring can be determined based on theirphenotype. |
If the genes are independently assorted, the dihybrid back-crosswould result in only 2 genotypes in the F1 generation. |
B and C |
Why do we map genes?
| To understand how genes interact with each other |
| Comparative genomics analysis |
| To determine the genotype of an organism |
| All of the above |
