Some mutations in the tra gene of Drosophila cause XX individuals to appear as males. In contrast, other mutations in tra cause XY flies to appear as females. Which of these
HINT: This problem involves an understanding of the Sxl regulatory pathway in Drosophila. The key to its solution is to review the functions of the normal TRA protein in Drosophila males and females.
To describe: The phenotype that would be expected if a mutation in tra gene results into null allele.
Introduction: The sex of Drosophila melanogaster is determined by the ratio of sex chromosomes to autosomes. The male Drosophila melanogaster is produced when the ratio of sex chromosomes to autosomes is 0.5. However, the female Drosophila melanogaster is produced when the ratio is 1.0.
Explanation of Solution
The tra gene is present in both male and female Drosophila melanogaster. The activity of tra gene is controlled by Sxl gene. In the absence of Sxl gene, the tra gene is spliced in a way that stop codon “UGA” remains present in the mature mRNA. This is observed in male Drosophila melanogaster as they lack the Sxl gene. The translation of mRNA with stop codon leads to the formation of a non-functional protein. However, in female Drosophila melanogaster, the Sxl gene promotes the removal of “UGA” from mRNA, and this forms a functional tra protein.
The mutation in the null allele leads to the formation of male Drosophila melanogaster. This is because the null gene has no effect on the Sxl gene. It does not activate the Sxl gene. The inactivated Sxl gene cannot remove the UGA gene from mRNA, and this leads to the formation of male Drosophila.
Thus, male Drosophila melanogaster would be expected if a mutation in the tra gene occurs in null allele.
To determine: The phenotype that would be expected if mutation produces a constitutively active tra gene.
Introduction: The ratio of sex chromosomes to autosomes is termed as chromosomal ratio. The chromosomal ratio is analyzed by three genes. These are sex-lethal (Sxl), transformer (tra), and doublesex (dsx). The activity of tra gene plays a crucial role in determining the sex of Drosophila.
Explanation of Solution
The activity of tra gene depends upon the Sxl gene. The active tra gene indicates that there is Sxl gene. The tra gene in the presence of the Sxl gene is able to remove the stop codon “UGA” from the mRNA. It produces a function tra protein. The active tra protein is found in female Drosophila. However, the male Drosophila has inactive tra protein. This indicates that female Drosophila would be expected if the mutation produces a constitutively active tra gene.
Thus, the phenotype that would be expected if mutation produces a constitutively active tra gene is a female Drosophila melanogaster.
Want to see more full solutions like this?
Chapter 18 Solutions
Concepts of Genetics (12th Edition)
- Identify the indicated structure. (Cucurbita stem l.s.) pit lenticel stomate tendril none of thesearrow_forwardIdentify the specific cell? (Zebrina leaf peel) vessel element sieve element companion cell tracheid guard cell subsidiary cell none of thesearrow_forwardWhat type of cells flank the opening on either side? (leaf x.s.) vessel elements sieve elements companion cells tracheids guard cells none of thesearrow_forward
- What specific cell is indicated. (Cucurbita stem I.s.) vessel element sieve element O companion cell tracheid guard cell none of thesearrow_forwardWhat specific cell is indicated? (Aristolochia stem x.s.) vessel element sieve element ○ companion cell O O O O O tracheid O guard cell none of thesearrow_forwardIdentify the tissue. parenchyma collenchyma sclerenchyma ○ xylem O phloem O none of thesearrow_forward
- Please answer q3arrow_forwardRespond to the following in a minimum of 175 words: How might CRISPR-Cas 9 be used in research or, eventually, therapeutically in patients? What are some potential ethical issues associated with using this technology? Do the advantages of using this technology outweigh the disadvantages (or vice versa)? Explain your position.arrow_forwardYou are studying the effect of directional selection on body height in three populations (graphs a, b, and c below). (a) What is the selection differential? Show your calculation. (2 pts) (b) Which population has the highest narrow sense heritability for height? Explain your answer. (2 pts) (c) If you examined the offspring in the next generation in each population, which population would have the highest mean height? Why? (2 pts) (a) Midoffspring height (average height of offspring) Short Short Short Short (c) Short (b) Short Tall Short Tall Short Short Tall Midparent height (average height of Mean of population = 65 inches Mean of breading parents = 70 inches Mean of population = 65 inches Mean of breading parents = 70 inches Mean of population = 65 inches Mean of breading parents = 70 inchesarrow_forward
- P You are studying a population of 100 flowers that has two alleles at a locus for flower color, blue (B) and green (G). There are 15 individuals with the BB genotype, 70 individuals with the BG genotype, and 15 individuals with the GG genotype. (a) What are the allele frequencies of B and G in the starting population? Show your calculations. (2 pts) (b) Is this population in Hardy-Weinberg equilibrium? Show your calculations. (3 pts) 12pt v Paragraph BIU UA AV & VT2V f CO Varrow_forwardIn a natural population of outbreeding plants, the variance of the total number of seeds per plant is 16. From the natural population, 20 plants are taken into the laboratory and developed into separate true-breeding lines by self- fertilization-with selection for high, low, or medium number of seeds-for 10 generations. The average variance in the tenth generation in each of the 20 sets is about equal and averages 5.8 across all the sets. Estimate the broad-sense heritability for seed number in this population. (4 pts) 12pt v Paragraph BIUA V V T² v B ① O wordsarrow_forwardIn a natural population of outbreeding plants, the variance of the total number of seeds per plant is 16. From the natural population, 20 plants are taken into the laboratory and developed into separate true-breeding lines by self- fertilization-with selection for high, low, or medium number of seeds-for 10 generations. The average variance in the tenth generation in each of the 20 sets is about equal and averages 5.8 across all the sets. Estimate the broad-sense heritability for seed number in this population. (4 pts) 12pt v Paragraph BI DI T² v ✓ B°arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage LearningBiology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage LearningBiology Today and Tomorrow without Physiology (Mi...BiologyISBN:9781305117396Author:Cecie Starr, Christine Evers, Lisa StarrPublisher:Cengage Learning
- Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxConcepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College