Concept explainers
a.
To determine:
The screening method that would enable to identify mutations in additional genes required for R7 determination.
Introduction:
A null mutation is a loss-of-function mutation. It can lead to complete loss of production of the mutated gene product or product that does not function properly. A null allele can be distinguished from deletion of an entire locus only from the observation of the
b.
To determine:
Whether it is possible to recover mutations in every gene required for R7 development with the above method.
Introduction:
The screens for loss-of-function mutants, which shows particular phenotypic defects can identify genes dedicated to a specific developmental pathway. In the development pathway, the gene will be expressed to give a specific outcome.
c.
To determine:
Whether the new mutations found in the screen are alleles of sev or boss.
Introduction:
The researchers now have the genome sequences for each model organism. It is possible to find the genes which are corresponding to any mutant phenotype.
d.
To determine:
The method to use a newly found one recessive mutant allele in a new mutagenesis screen to find additional alleles of this gene and the reason due to which these additional mutant alleles are required.
Introduction:
The screen in which any organisms with pre-existing phenotype are used to identify mutations that can enhance or suppress this phenotype is termed as a modifier screen. This method can reveal missing pathway members, original genetic interactions, and new drug targets.
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Genetics: From Genes to Genomes
- You are interested in studying position effect variegation in Drosophila using the chromosome depicted below: Deactivation of the w+ gene gives a white eye phenotype and deactivation of the rst+ gene gives a rough eye phenotype; the normal phenotypes are red and smooth. Because the rst+ and w+ genes have now been placed close to a heterochromatic domain, some sections (or sectors) of the eye display mutant phenotypes due to gene deactivation while others display the normal, wild type phenotype. Which phenotype would you not expect to see rst w Inverted X chromosome white smooth eye sectors white rough eye sectors red smooth eye sectors red rough eye sectorsarrow_forwardYou are a developmental geneticist studying flowering time variation in Arabidopsis. You perform a mutagenesis screen to identify mutants in the photoperiod pathway. You conduct the screen and find two different plants that show the same mutant phenotype. You then use a complementation test. What is the predicted outcome of this test if both phenotypes are caused by mutations in separate genes? recover the wild type phenotype overexpress the gene O recover the mutant phenotypearrow_forwardThe Drosophila gene Sex lethal (Sxl) is deserving of itsname. Certain alleles have no effect on XY animals butcause XX animals to die early in development. Other alleles have no effect on XX animals but cause XY animals to die early in development. Thus, some Sxl allelesare lethal to females, while others are lethal to males.a. Would you expect a null mutation in Sxl to causelethality in males or in females? b. Why do Sxl alleles of either type cause lethality ina specific sex?The gene transformer (tra) gets its name from sexualtransformation, as some tra alleles can change XXanimals into morphological males, while other traalleles can change XY animals into morphologicalfemales.c. Which of these sex transformations would becaused by null alleles of tra and which would becaused by constitutively active alleles of tra?d. In contrast with Sxl, null tra mutations do notcause lethality either in XX or in XY animals.However, the Sxl protein regulates the productionof the Tra protein. Why…arrow_forward
- I. Male Drosophila from a true-breeding wild-typestock were irradiated with X-rays and then mated withfemales from a true-breeding stock carrying the following recessive mutations on the X chromosome:yellow body (y), crossveinless wings (cv), cut wings(ct), singed bristles (sn), and miniature wings (m).These markers are known to map in the order:y - cv - ct - sn - mMost of the female progeny of this cross were phenotypically wild type, but one female exhibited ct and snphenotypes. When this exceptional ct sn female wasmated with a male from the true-breeding wild-typestock, twice as many females as males appearedamong the progeny.a. What is the nature of the X-ray-induced mutationpresent in the exceptional female?b. Draw the X chromosomes present in the exceptional ct sn female as they would appear duringpairing in meiosis.c. What phenotypic classes would you expect to seeamong the progeny produced by mating the exceptional ct sn female with a normal male from a truebreeding wild-type…arrow_forwardPlease indicate true or false about the statements regarding flies with the following genotype: Hs>FLP; FRT82B, Actin-GFP, e2f1i2. e2f1i2 is a mutation and that both this mutation and Actin-GFP reporter are located on the same chromosomal arm as FRT82B. This line can be crossed with w; FRT82B to create mosaic clones in wing discs after heat shock. Cells with higher GFP intensity correspond to the WT and the lower intensity GFP-positive cells are heterozygous. When this line was crossed with w; FRT82B males, no mosaic clones will be generated in wing discs even after heat shock. All the cells in wing discs will have similar level of GFP expression. This line can be crossed with w; FRT82B to create mosaic clones in wing discs after heat shock. Cells with no GFP expression can be used to identify the WT cells. This line can be crossed with w; FRT82B to create mosaic clones in wing discs after heat shock. Cells with no GFP…arrow_forwardMutations in the HPRT1 gene in humans result in atleast two clinical syndromes. Consult OMIM (www.omim.org) by querying HPRT1; you will only needto look briefly at the top three hits (files #300322,300323, and 308000).a. What is the full name of the HPRT1 enzyme?b. On which chromosome is the HPRT1 gene located?c. Mutations in HPRT1 are associated with two different syndromes. What are these syndromes? Foreach, answer the following questions: (i) What arethe symptoms associated with the syndrome? (ii) Isthe mutant allele that causes the syndrome dominant, recessive, codominant, or incompletely dominant with respect to the normal allele, or do specialconditions apply? (iii) Is the syndrome associatedwith a loss-of-function or a gain-of-function disease allele? (iv) Does the syndrome display allelicheterogeneity? (v) Does the syndrome display locus heterogeneity? (Note: You do not need to understand everything in the OMIM entries to answerthese questions.)arrow_forward
- In Drosophila, the wildtype eye color is black. In the laboratory, you screened for mutants than when homozygous results to different eye phenotypes. You found two mutants 1) red eye (re) and 2) white eye (we). You performed a complementation test and the resulting eye phenotype was gray. From the results, what conclusion can you make? A. There was complementation, thus the two mutations are alleles of different genes. B. There was no complementation, thus the two mutations are alleles of different genes. C. The two mutations failed to complement, thus they are alleles of different genes. D. The two mutations failed to complement, thus they are alleles of the same genes. E. The two mutations complemented, hence they are likely controlled by different genes.arrow_forwardImagine a scenario in which prenatal testing of a human female fetus indicates that the baby will have a normal XX karyotype but is heterozygous for a mutation that inactivates the Xist promoter. Allele “Xr” represents the mutated version of the Xist promoter, and “XR” represents the normal version of the Xist promoter. How will this mutation affect the process of X inactivation?A. "X inactivation will still be random with both mutant and normal X chromosomes being randomly inactivated" B. "The chromosome with the mutant Xist promoter will always be active" C. "The chromosome with the mutant Xist promoter will always be inactivated" D. "The normal X chromosome (no mutation in Xist) will always be active" E. "The normal X chromosome (no mutation in Xist) will always be inactivated"arrow_forwardConcerning the Tools of Genetics Box Analysis ofCell-Cycle Mutants in Yeast:a. Describe how you would use replica plating ofmutagenized, haploid yeast cells to identifytemperature-sensitive (ts) mutations in essentialgenes needed for yeast growth and survival.b. Among the many ts mutations you found in part(a), how would you distinguish mutations in genesneeded for cell-cycle progression from those ingenes needed for other aspects of the life of yeasts?c. If you had a large collection of yeast cell-cyclemutants, how would you determine which of themutations are in the same gene and which are indifferent genes?arrow_forward
- Leber Congenital Amaurosis (LCA) causes progressive vision loss due to defects in the gene that encodes RPE65 isomerase. Affected individuals are homozygous recessive for mutant alleles of the RPE65 gene. You are trying to determine the molecular nature of the mutations in three individuals with LCA. For ease of analysis, you may assume that each individual is homozygous for the same mutant allele (though the three individuals have different mutations than each other). You use the polymerase chain reaction to amplify DNA from each patient and you determine the sequence of the DNA and compare it to unaffected individuals. You identify the following differences. Note that the non-template strand of DNA is given and the changes are highlighted using red boldface. You can assume that the sequences are in the first reading frame (eg. the first three nucleotides of each sequence is a codon). The coding region of the gene is 1602 bp and the position of the sequences shown below is…arrow_forwardIn Figure 6-1,a. what do the yellow stars represent?b. explain in your own words why the heterozygote isfunctionally wild typearrow_forwardDiseases which are caused by recessive variants in loci located on the X chromosome affect females and males differently. How? Lifestyle choices and environmental exposures can also shape up a disease, how? (Discuss from the aspects of molecular biology) c. In a designed Microarray experiment, in which the probes for the control cell’s transcripts are labelled with green dye and probes for cancer cell’s transcripts are labelled with red dye. What color intensities will you expect in the spots of oncogenes and tumor suppressor genes?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning