3) What genotypes and phenotypes do you expect as a result of the yellow body, medium bristled X yellow body, medium bristled cross, and in what frequencies will they occur? Show your work and include only the live offspring in these results.
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- The Y allele in Drosophila is a dominant for yellow body color and the y allele is recessive for brown body color. The Y allele is also a recessive lethal allele. At a second locus, the allele B for long bristles is incompletely dominant to the allele b for short bristles. Follow the steps below to predict the genotypes and phenotypes of the LIVE progeny from the following cross: yellow body, medium bristled X yellow body, medium bristled. 1) What is the cross? 2) Break this down to the single gene level by answering the following questions: a. What is the ratio of yellow to brown offspring? Show the Punnett square, but include only the live offspring in the final ratios, and place a box around this final ratio: b. What is the ratio of bristle lengths? Show the Punnett square:In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn and ct+ is crossed to a sn+ct male. The F1 flies are interbred. The F2 males are distributed as follows: genotype number sn ct 15 sn ct+ 34 sn+ ct 33 sn+ct+ 18 What is the map distance between sn and ct?In Drosophila, singed bristles (sn) and cut wings (ct) are both caused by recessive, X-linked alleles. The wild type alleles (sn+ and ct+) are responsible for straight bristles and intact wings, respectively. A female homozygous for sn+ and ct+ is crossed to a sn ct male. The F1 flies are interbred. The F2 males are distributed as follows sn ct 36 sn ct+ 13 sn+ ct 12 sn+ ct+ 39 What is the map distance between sn and ct?
- In the following cross, imagine that you have a female fly that has two Xs and one Y due to a nondisjunction event in her mother's germ cells. Draw out what the possible gametes are for both the female and the male and also a Punnett square showing the genotypes, phenotypes, and sex of the possible flies as a result of this cross. You do not need to provide the probabilities of each of these. Red-eyed wi C Ở Red-eyed wt XX Y X Y MeiosisMendelian ratios are modified in crosses involving autotetraploids.Assume that one plant expresses the dominant trait greenseeds and is homozygous (WWWW). This plant is crossed to onewith white seeds that is also homozygous (wwww). If only onedominant allele is sufficient to produce green seeds, predict theF1 and F2 results of such a cross. Assume that synapsis betweenchromosome pairs is random during meiosis.In tomatoes, dwarf (d) is recessive to tall (D) and opaque (light-green) leaves (op) are recessive to green leaves (Op). The loci that determine height and leaf color are linked and separated by a distance of 7 m.u. For each of the following crosses, determine the phenotypes and proportions of progeny produced.
- A cross was performed using Drosophila melanogaster involving a female known to be heterozygous for both ebony body and sepia eyes and a male known to be homozygous wild type male. The resulting progeny were allowed to mate with one another to produce the data set. Three repetitions of the experiment were conducted. The following data were produced from the crosses. Test these data to determine if they are significantly different from the expected phenotypic ratio. Use the 5% level of significance. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject for each experiment. Wild eye Wild body – 112, Wild eye Ebony body – 40, Sepia eye Wild body – 35, Sepia eye Ebony body – 11In goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We’ll use the symbol Bb for the beard allele and B+ for the beardless allele. Another independently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Give the phenotypes and their expected proportions for the following crosses. Q. B+ Bb Ww male × Bb Bb ww femaleIn Drosophila, the sepia mutation (se, chromosome 3, position 26) results in dark brown eyes, while cinnabar (cn, chromosome 2, position 57.5) results in bright orange-red eyes. True breeding, wild type females are mated with true breeding males homozygous recessive for both traits. Using Drosophila notation, diagram the P1 and F1 crosses. P1 F1 Fill in the chart with phenotypic ratios that would be expected in the F2 generation. Use the space provided to show your work. Phenotype Females Males Overall (♀and ♂) =1 =1
- Short hair (S) in rabbits is dominant over long hair (s). The following crosses are carried out, producing the progeny shown. Give all possible genotypes of the parents in each cross.A standard three-point mapping is conducted for recessive mutations in autosomal genes purple eye (pr), curved wing ( c) and black body (b). Their wild type alleles are also used for genetic mapping. An F1 Drosophila female heterozygous for purple eye (pr), curved wing (c) and black (b) is crossed to a triply homozygous mutant male. The observed numbers and phenotypes of the offspring are as follows: 360 pr c b 380 pr+ c+ b+ 104 pr c+ b 96 pr+ c b+ 30 pr c b+ 20 pr+ c+ b 6 pr c+ b+ 4 pr+ c b PROVIDE THE FOLLOWING: A) State the order of genes on this chromosome. B) Calculate map distances between the gene pairs: pr-c, pr-b, c-b. Show calculations, state the number of map units and which gene pairs they refer to.In goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We’ll use the symbol Bb for the beard allele and B+ for the beardless allele. Another independently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Give the phenotypes and their expected proportions for the following crosses. a. B+ Bb Ww male × B+ Bb Ww female b. B+ Bb Ww male × B+ Bb ww female c. B+ B+ Ww male × Bb Bb Ww female d. B+ Bb Ww male × Bb Bb ww female