Question

In: Biology

F1 Wild Type : Red Eyes and Long Wings & Mutants: White Eyes and Short Wings...

F1 Wild Type : Red Eyes and Long Wings & Mutants: White Eyes and Short Wings

Mutant Female x Wild Type Male
Wild Type Wing & Red Eyes Females 121
Mutant Wing & White Eyes Males 95
Wild Type Female x Mutate Male
Wild Type Wing & Red Eyes Females 43
Wild Type Wing & Red Eyes Males 45

F2

Blue Label Red Label
F1 Female x Mutant Male F1 Female x Mutant Male
Wild Type Wing & Red Eyes Female 91 Wild Type Wing & Red Eyes Female 42
Wild Type Wing & Red Eyes Males 49 Wild Type Wing & Red Eyes Males 43
Short Wing & Red Eyes Female 0 Short Wing & Red Eyes Female 15
Short Wing & Red Eyes Males 0 Short Wing & Red Eyes Males 14
Wild Type Wing & White Eyes Female 10 Wild Type Wing & White Eyes Female 31
Wild Type Wing & White Eyes Males 5 Wild Type Wing & White Eyes Males 20
Short Wing & White Eyes Female 23 Short Wing & White Eyes Female 42
Short Wing & White Eyes Males 31 Short Wing & White Eyes Males 42
209 249
F1 Female x F1 Males F1 Female x F1 Males
Wild Type Wing & Red Eyes Female 202 Wild Type Wing & Red Eyes Female 51
Wild Type Wing & Red Eyes Males 72 Wild Type Wing & Red Eyes Males 50
Short Wing & Red Eyes Female 3 Short Wing & Red Eyes Female 11
Short Wing & Red Eyes Males 13 Short Wing & Red Eyes Males 12
Wild Type Wing & White Eyes Female 16 Wild Type Wing & White Eyes Female 15
Wild Type Wing & White Eyes Males 6 Wild Type Wing & White Eyes Males 27
Short Wing & White Eyes Female 11 Short Wing & White Eyes Female 36
Short Wing & White Eyes Males 62 Short Wing & White Eyes Males 32
385 234

What is the mode of transmission for each trait, autosomal recessive, autosomal dominant, or sex-linked recessive or sex-linked dominate? What is the recombination frequency?

Solutions

Expert Solution

In mutant female X wild type male, all the female offsprings are wildtype, while all the male offsprings are mutants for both traits. The F2 generation in the red label in case of back crossing (with mutant male) and F1 crossing are similar. This confirms that F1 males are mutants for both traits.

But in wild type female X mutant male, all the offsprings are wildtype and there is almost equal number of males and females. In the F2 generation, we can observe that there are more females with long wing and there are more males with short wing. Therefore, the trait for length of wings is sex-linked recessive. But there is no such difference between females and males in the case of eye color. Therefore the trait for eye color is autosomal recessive.

SInce both genes are linked (which is evident from the data), let us calculate the recombination frequency.

Recombination frequency is a measure of genetic linkage and is calculated as the percentage of recombinants present in a generation.

In the F2 generation (F1 x F1),

Total number of progeny (both red and blue label) =

Total number of recombinants =

Therefore,


Related Solutions

In Drosophila, red eyes (E) are dominant over white (e) and normal wings (W) are dominant...
In Drosophila, red eyes (E) are dominant over white (e) and normal wings (W) are dominant over shriveled wings (w). Use the addition and multiplication rules to calculate the phenotypic ratios of the cross of Eeww and EeWw. 1. Cross of Ee x Ee: 2. Cross of ww x Ww: 3. Red eyes and normal wings: 4. Red eyes and shriveled wings: 5. White eyes and normal wings: 6. White eyes and shriveled wings:
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...
32) Normal (wild-type) Drosophila eyes are red. In a cross in Drosophila involving the X-linked recessive...
32) Normal (wild-type) Drosophila eyes are red. In a cross in Drosophila involving the X-linked recessive eye mutation white (resulting in a white eye) and the autosomal recessive eye mutation sepia (resulting in a dark eye), predict the sex and color of the eyes of the offspring from crossing true-breeding parents for parts A) and B) below. White is epistatic to the expression of sepia (i.e., if an eye is white, it cannot also be sepia; i.e., white masks the...
A female fruit fly with the recessive mutant phenotype of white eyes and miniature wings is...
A female fruit fly with the recessive mutant phenotype of white eyes and miniature wings is mated with a male possessing the wild-type phenotype of red eyes and normal wings. Among the F1s, all the females are wild-type while all the males exhibit the mutant phenotype. The F2s resulting from a test cross exhibit predominantly (63%) parental phenotypes. How do you explain these results?
In the tubular flowers of foxgloves, wild-type coloration is red while a mutation called white produces...
In the tubular flowers of foxgloves, wild-type coloration is red while a mutation called white produces white flowers. Another mutation, called peloria, causes the flowers at the apex of the stem to be huge. Yet another mutation, called dwarf, affects stem length. You crossed a red-flowered plant (gene symbol, r) to a plant that is dwarf (gene symbol, d) and peloria (gene symbol, p). All of the F1 plants are tall with red, normal stem flowers. You cross an F1...
Offspring of fruit flies may have red or brown bodies and normal wings or short wings....
Offspring of fruit flies may have red or brown bodies and normal wings or short wings. Genetic theory predicts that these traits will appear in the ratio​ 9:3:3:1 (9 red​, ​normal: 3 red​, ​short: 3 brown​, ​normal: 1 brown​, ​short). A researcher checks 100 such flies and finds the distribution of the traits to be 55​, 20​, 14​, and 11​, respectively. Compute the​ chi-square statistic. x²= critical value? P-value? Suppose you double the amount of flies to 200 and the...
In Drosophila the recessive alleles for white eyes is X-linked (normal eye color is red) and...
In Drosophila the recessive alleles for white eyes is X-linked (normal eye color is red) and the recessive allele for vestigial wings is autosomal. A female heterozygous for both of these traits is crossed with a white-eyed vestigial male. What proportion of the FEMALE progeny will be white-eyed with vestigial wings.? (a) 100 %    (b) 75 %       (c) 50 %     (d) 25 %     (e) 0 %
a. Your dream is to find durum seeds that are white in color (wild type is...
a. Your dream is to find durum seeds that are white in color (wild type is amber or yellow) and so you mutagenize thousands of amber seeds and identify 6 plants segregating for white seeds I. How would you identify true breeding white lines? II. Determine whether the 6 individual white lines contained mutations in the same gene or different genes? b. The following two genotypes are crossed: Aa Bb Cc dd Ee × Aa bb Cc Dd Ee. What...
A wild legume with white flowers and long pods is crossed to one with purple flowers...
A wild legume with white flowers and long pods is crossed to one with purple flowers and short pods. The F1 offspring are allowed to self-fertilize, and the F2 generation has 301 long purple, 99 short purple, 612 long pink, 195 short pink, 295 long white and 98 short white plants. How are these traits being inherited? Inheritance of these traits can best be explained by one gene with 3 different alleles.   Inheritance of these traits can best be explained...
Cinnabar eyes (cn) and reduced bristles (rd) are autosomal recessive characters in Drosophila. A homozygous wild-type...
Cinnabar eyes (cn) and reduced bristles (rd) are autosomal recessive characters in Drosophila. A homozygous wild-type female was crossed to a reduced, cinnabar male, and the F1 males were then crossed to the F1 females to obtain the F2. Of the 400 F2 offspring obtained, 292 were wild type, 9 were cinnabar, 7 were reduced, and 92 were reduced and cinnabar. Briefly explain (2 sentences) these results and estimate the distance between the cn and rd loci.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT