Question

In: Biology

Imagine that you as a fruit fly researcher cross true breeding, wild-type flies to true breeding,...

Imagine that you as a fruit fly researcher cross true breeding, wild-type flies to true breeding, eyeless flies in the P generation. The resulting F1 generation is made up of 100% wild-type flies with normal eyes. Based on Mendelian genetics, what phenotype ratio would you expect to see for a F1 monohybrid cross? How would you go about determining the genotype ratio in the F2 generation?
In cactus, the relationship between Gene S and Gene N is known to be epistatic. Gene S controls the sharpness of spines in a type of cactus. Cacti with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cacti have dull spines. At the same time, the second gene, N, determines whether or not cacti have spines. Homozygous recessive nn cacti have no spines at all. Briefly explain why this genetic interaction is considered epistasis. Which gene is considered epistatic to the other?
In Drosophila melanogaster, vestigial wings are determined by a recessive allele of a gene that is linked to a gene with a recessive allele that determines black body color. T. H. Morgan crossed black-bodied, normal-winged females and gray-bodied, vestigial-winged males. The F1 were all gray bodied, normal winged. The F1 females were crossed to homozygous recessive males to produce testcross progeny. Morgan calculated the map distance to be 17 map units. Which of the phenotypes used in the experiment would be considered parental? Which phenotypes would be recombinants? What would be the percentages of the resulting phenotypes of the testcross progeny?
The most commonly occurring mutation in people with cystic fibrosis is a deletion of a single codon. What is the result of this type of mutation at the protein level? Which class of mutation would have an even more severe effect on the protein?

Please explain!

Expert Solution

1) A monohybrid cross for is performed between a Normal eyed Wild type (AA) and true breeding eyeless flies (aa).

Parental cross : AA (Wild Type) x aa (Eyeless)

Gametes: A A a a

F1 Progeny

Male/Female	A	A
a	Aa	Aa
a	Aa	Aa

Genotype: All are heterozygotes

Phenotype: All all Normal Eyed

2) In a cross between two F1 generations.

F2 cross : F1 x F1

Gametes: A a A a

F2 progeny:

Male/Female	A	a
A	AA	Aa
a	Aa	aa

Genotype: 1AA : 2Aa : 1aa

Phenotype: 3 Normal eyed : 1 eyeless

3) The Following conclusion can be made from the question

S_ - Sharp Spines

ss - Dull spines

N_ - Spines

nn- No spines

The Phenomenon observed here is know and Recessive epistasis (9:3:4). Recessive epistasis occurs when the recessive allele at one loci or gene masks or modifies the effect alleles at other loci or gene. Here the function of the gene products of both the alleles occur in series, that is a precursor molecule X is converted to an intermediate precursor A by S_ genotype and this intermediate A precursor is converted to final phenotype by N_ genotype.

4) The recessive allele ss is epistatic over the expression of N allele.

gladiator answered 2 years ago

A female fruit fly with singed bristles was mated with a male from a true-breeding wild-type...

A female fruit fly with singed bristles was mated with a male from a true-breeding wild-type stock with long bristles. All of the F1 females had wild-type long bristles and all of the F1 males had singed bristles. If the F1 flies are intercrossed, the expected ratio of long to singed bristles in the F2 flies is Multiple Choice 3:1 in both sexes. 3:1 in females, while all the males will have singed bristles. 1:1 in both sexes. 1:1 in...

Problem 2 You cross a true-breeding yellow-bodied, smooth-winged female fly with a true-breeding red-bodied, crinkle-winged male....

Problem 2 You cross a true-breeding yellow-bodied, smooth-winged female fly with a true-breeding red-bodied, crinkle-winged male. The red body phenotype is dominant to the yellow body phenotype and smooth wings are dominant to crinkled wings. Use B or b for body color alleles, and W or w for wing surface alleles. a) What are the genotypes of the P generation flies? b) What will be the genotype(s) and phenotype(s) of the F1 offspring? c) You discover that the genes for...

In the fruit fly Drosophila melanogaster, wild type eye color is a brick red color and...

In the fruit fly Drosophila melanogaster, wild type eye color is a brick red color and is produced by the action of two genes, one producing red color, the other brown. Vermilion is an X-linked recessive mutation that results in a bright red eye. Brown is an autosomal recessive mutation that results in a brown eye. Flies carrying both the vermilion and brown mutations produce no pigment and have white eyes. True breeding Vermilion eyed females are crossed to brown...

Purebred lines of fruit flies with wild type (tan) body color and stubby bristles are mated...

Purebred lines of fruit flies with wild type (tan) body color and stubby bristles are mated to flies with ebony bodies and normal bristles. The resulting F1 offspring all have a normal wild-type body color and stubby bristles. The F1 flies are crossed with flies recessive for both traits (i.e. ebony bodies and normal bristles). a. What phenotypes and ratios of phenotypes would you expect to find among the offspring produced by crossing the F1 with flies recessive for both...

In fruit flies, bristle shape is controlled by an X-linked gene. The dominant wild-type allele (+)...

In fruit flies, bristle shape is controlled by an X-linked gene. The dominant wild-type allele (+) results in normal bristles, while the recessive allele (sn) results in short 'singed' bristles. A normal-bristled female offspring of a male with singed bristles is crossed with a normal-bristled male. If we consider only the progeny that have normal bristles, what is the ratio of females : males among these? 2:3 1:3 3:1 1:1 2:1 3:2 1:2

White eye is a recessive X-linked mutant in fruit flies. A wild type male is mated...

White eye is a recessive X-linked mutant in fruit flies. A wild type male is mated to a white-eyed female. What is the probability that an F1 son will be white eyed? A. 0 B. 0.25 C. 0.5 D. 0.75 E. 1.0

A fly was found that had two mutations, Q&Z. Reciprocal crosses between wild type flies and...

A fly was found that had two mutations, Q&Z. Reciprocal crosses between wild type flies and Q&Z mutants gave all Q&Z mutant F1 progeny in all cases. These results suggest that the Q&Z mutations have an autosomal dominant inheritance pattern. The genes X, Q, and Z appear to be linked. To map the genes the following crosses were done. A homozygous Q, Z mutant female was crossed to a homozygous X mutant males shown in the diagram below. X, Q,...

Long wings in fruit flies are dominant to vestigial wings. A long winged fruit fly is...

Long wings in fruit flies are dominant to vestigial wings. A long winged fruit fly is crossed with a vestigial winged fly. What are the genotypes of the parents and offspring if the cross results in 72 long winged and 68 vestigial winged offspring.

Wild fruit flies have red eyes. A recessive mutation produces white-eyed individuals. A researcher wants to...

Wild fruit flies have red eyes. A recessive mutation produces white-eyed individuals. A researcher wants to assess the proportion of heterozygous individuals. A heterozygous red-eyed fly can be identified through its off-spring. When crossed with a white-eyed fly it will have a mixed progeny. A random sample of 100 red-eyed fruit flies was taken. Each was crossed with a white- eyed fly. Of the sample flies, 12 were shown to be heterozygous because they produced mixed progeny. a) Check this...

A true breeding red eyed fly with long bristles was crossed to a true breeding white...

A true breeding red eyed fly with long bristles was crossed to a true breeding white eyed fly with stubble bristles. All F1 progeny are red eyed with stubble bristles. An F1 female is crossed to a tester male and the following numbers of progeny are observed: 56 red eyed, stubble bristles; 72 red eyed, long bristles; 61 white eyed stubble bristles; and 68 white eyed, long bristles. You are asked to perform a chi-squared test for this result. 1....