In Your Experiments With Drosophila You Found That The Wild
In Your Experiments With Drosophila You Found That The Wildtype Allel
In your experiments with Drosophila, you observed that the wildtype allele coding for grey adult insects can be mutated, resulting in black-colored mutants. Similarly, the population contains alleles for vestigial wings instead of the wildtype normal wings, with both mutations being recessive. You aim to determine whether the two genes for body colour and wings are linked. Your preliminary crosses involve establishing a large number of individuals for testcrosses.
9a. Which two genotypes will you use for the testcross? (Hint: One of the two genotypes should be a heterozygous dihybrid.) Use the common wildtype notation (e.g., “+” for wildtype alleles, “b” for black mutation, “vg” for vestigial wing mutation). Disregard gender.
Answer:
The testcross should involve crossing a heterozygous dihybrid individual with a double recessive individual. The heterozygous dihybrid genotype in wildtype notation is /+ + / v g, where “+” symbolizes the wildtype alleles for body colour and wings, and “v g” denote the two recessive mutations (black body and vestigial wings). The tester (recessive) individual must be homozygous recessive for both traits: b b v g v g.
Thus, the two genotypes for the testcross are:
- /+ + / v g (heterozygous dihybrid, wildtype alleles)
- b b v g v g (homozygous recessive for both genes)
9b. Indicate the two phenotypes of your testcross.
Answer:
The testcross will produce offspring with four possible phenotype combinations:
- Grey body with normal wings (wildtype phenotype)
- Black body with normal wings
- Grey body with vestigial wings
- Black body with vestigial wings
9c. Indicate the genotypes of the progeny (next generation) of the testcross by drawing a Punnett square.
Answer:
Set up the Punnett square with the gametes:
| | + / v g | + / v g |
|-----------|-----------|-----------|
| b / v g | +b v g | +b v g |
| b / v g | +b v g | +b v g |
Actually, for clarity, the gametes from each parent:
- Heterozygous parent: + v g × + v g produces gametes: +, +, v, g (but more accurately, the heterozygous parent: +/ v g, which produces gametes: +, v, g, + v g). For simplifying, we consider the typical dihybrid segregation.
The Punnett square reveals:
| | + / v g | + / v g |
|---------------------|---------|---------|
| b / v g | + b / v g | + b / v g |
| b / v g | + b / v g | + b / v g |
But considering the actual Mendelian segregation, the progeny genotypes will include:
- +/ v g / + / v g (wildtype phenotype)
- b / + / v g (black body, wildtype wings)
- +/ + / v g (grey body, vestigial wings)
- b / b / v g (black body, vestigial wings)
In the simplified dihybrid cross, expectations based on independent assortment would produce a 1:1:1:1 ratio of these four classes.
9d. Assuming a total number of 4400 flies for the progeny generation, what are your expectations for the ratio of genotypes if the two genes are linked?
Answer:
If the two genes are linked, they tend to be inherited together more often than expected by independent assortment. In this case, the progeny would mainly consist of parental-type phenotypes, with fewer recombinant types.
Assuming no crossing-over occurs (complete linkage), the expected genotypes would only be parental types:
- Grey body with normal wings (from the original heterozygote parent)
- Black body with vestigial wings
Hence, the expected ratios are approximately:
- Parental phenotype 1 (grey + normal): 50% (2200 flies)
- Parental phenotype 2 (black + vestigial): 50% (2200 flies)
Recombinants would be rare or absent unless crossing-over occurs.
9e. The two genes are not linked.
Answer:
If the genes are not linked (assorting independently), the progeny genotypes would follow Mendel’s law of independent assortment, resulting in a 1:1:1:1 ratio among the four phenotypic classes:
- Grey + normal wings
- Black + normal wings
- Grey + vestigial wings
- Black + vestigial wings
This would mean approximately 1100 flies for each class out of 4400 total.
9f. Finally, if the two genes are not linked according to your results, would you expect that they lie on different chromosomes? Explain your answer.
Answer:
Yes, if the genes are unlinked (segregate independently), they are most likely located on different chromosomes. Mendelian principles suggest that genes on separate chromosomes assort independently during meiosis, leading to the 1:1:1:1 phenotypic ratio observed in the progeny. Conversely, linked genes tend to stay together, resulting in non-independent inheritance patterns. The lack of significant deviation from the 1:1:1:1 ratio in experimental data implies the genes are on different chromosomes or sufficiently far apart on the same chromosome to undergo crossing-over frequently enough to resemble independent assortment (if any recombinants are observed at expected frequencies).
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Paper For Above instruction
The investigation of gene linkage in Drosophila melanogaster provides crucial insights into the principles of inheritance, particularly Mendel's laws and genetic linkage. Understanding whether specific genes are linked—or inherited together—helps elucidate the physical arrangement of genes on chromosomes. This understanding is essential for mapping genes and exploring chromosomal behavior during meiosis.
In the described experiment, two traits are studied: body color (wildtype grey vs. mutant black) and wing morphology (wildtype normal wings vs. vestigial wings). Both mutations are recessive. The main goal is to determine whether these genes are linked or assort independently. To achieve this, a testcross is performed involving a heterozygous dihybrid organism and a homozygous recessive individual.
The chosen genotypes for the testcross are a heterozygous dihybrid /+ + / v g, and a double recessive b b v g v g. This pairing allows for assessment of all possible phenotypic combinations among offspring, making it possible to infer linkage. The phenotypes produced in the offspring include grey body with normal wings, black body with normal wings, grey body with vestigial wings, and black body with vestigial wings. These combinations reflect the inheritance patterns resulting from whether the two genes are linked or assort independently.
Drawing a Punnett square illustrates the expected genotype combinations. If genes are linked, the progeny will predominantly display parental types, with a minimal proportion of recombinants unless crossing-over occurs. In such a case, approximately 2200 out of 4400 flies would have parental phenotype combinations, providing evidence of linkage. Conversely, if the genes are unlinked, the progeny will adhere to a Mendelian 1:1:1:1 ratio, with about 1100 flies exhibiting each phenotype, indicating independent assortment.
The experimental outcomes serve to clarify whether the two genes are on the same chromosome. If the observed ratios support linkage, then the genes are likely located close together on a single chromosome. If the ratios follow independent assortment, they are situated on different chromosomes. The presence of linkage has profound implications in genetics, facilitating gene mapping and understanding chromosomal structure. Thus, this experiment underscores the importance of genetic linkage in inheritance and the mechanisms governing how genes are transmitted during meiosis.
In conclusion, evaluating the segregation ratios of phenotypes in progeny allows geneticists to determine whether genes are linked or assort independently. Such knowledge illuminates the complex architecture of genomes and advances our comprehension of heredity in model organisms like Drosophila. These principles are foundational to modern genetics and have broad applications in fields ranging from evolutionary biology to medical genetics.
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