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- 4e. You also study the expression of 3 different mutants for this gene. For each mutant answer the following: Does this mutation change the sequence of the protein produced? Why or why not? If it does change the sequence of protein be sure to write out the new sequence. If it does not change the protein sequence, what effect (if any) would you expect it to have on expression of the gene? 1 20 ORI 40 60 5'..TTCGAGCTCTCGTCGTCGAGATACGCGATGATATTACTGGTAATATGGGGATGCACTATC...3’ 3'...AAGCTCGAGAGCAGCAGCTCTATGCGCTACTATAATGACCATTATACCCCTACGTGATAG...5’ promoter i. Mutant A has a single base pair substitution with the T/A being replaced with C/G base pair at position 35 (position denoted by the * in the sequence above). ii. Mutant B has a 2 G/C pairs inserted between position 19 and 20 (position denoted by the ^ in the sequence above).2. Null mutations are valuable genetic resources becausethey allow a researcher to determine what happens to anorganism in the complete absence of a particular protein. However, it is often not a trivial matter to determinewhether a mutation represents the null state of the gene.a. Geneticists sometimes use the following test forthe nullness of an allele in a diploid organism: If theabnormal phenotype seen in a homozygote for theallele is identical to that seen in a heterozygote(where one chromosome carries the allele in question and the homologous chromosome is known tobe completely deleted for the gene) then the alleleis null. What is the underlying rationale for thistest? What limitations might there be in interpreting such a result?6. Suppose a particular gene is required for early development and also later for development of a particulartissue, such as the adult nervous system. By generating a homozygous mutant clone in that tissue of a heterozygote, researchers can circumvent the lethalitythat would result if the entire animal is homozygousfor a loss-of-function mutation in that gene.A technique called MARCM (Mosaic Analysiswith a Repressible Cell Marker) was developed to enable Drosophila geneticists to generate homozygousmutant cell clones that are marked by the presence of areporter protein such as GFP. Marker expression enables the investigator to observe clearly the mutantphenotype within a clone of mutant cells. This technique relies on a yeast protein called Gal80 that is anegative regulator of the Gal4 protein described previously in Solved Problem II. Gal80 binds to Gal4 andprevents it from activating transcription. The idea ofMARCM is that Gal4/UASG-driven GFP expression isblocked by Gal80 throughout…
- 1B. What type of mutation occurred in the gene sequence that resulted in white-fur bears?1. A monogenic disease is a disease caused by a mutation in a single gene. For instance, sickle-cell anemia is caused by a mutation in the HBB gene, which codes for the B- globin chain of hemoglobin. The beginning of HBB is shown here: 5'-ATGGTGCACCTGACTCCTGAGGAGAAGTCTGCCGTTACT...-3' A. Translate this HBB sequence into an amino acid sequence. B. In terms of amino acids, what is the result of the sickle cell mutation, wherein the bolded red A is changed to a T? This single mutation causes hemoglobin to aggregate, causing red blood cells to deform into a sickle-like shape rather than the normal “biconcave disk" shape. C. What would happen if the bolded blue A were mutated to at T? (This is hypothetical; it's not a mutation found in sickle-cell disease.)3. What Is heredity? O A. Certain tralts are always expressed in an organism B. The process by which genetic tralts are passed from parents to thelr offspring O C. The inheritance of a particular tralt does not affect the inheritance of another trait O D. A parent's genes separate, and only one half of the genes will then be passed on to the offspring 05:45:53
- 1. How is PKD inherited? What gene is responsible for the expression of PK enzyme ?. The production of pigment in the outer layer of seedsof corn requires each of the three independently assorting genes A, C, and R to be represented by at leastone dominant allele, as specified in Problem 64. Thedominant allele Pr of a fourth independently assortinggene is required to convert the biochemical precursorinto a purple pigment, and its recessive allele pr makesthe pigment red. Plants that do not produce pigmenthave yellow seeds. Consider a cross of a strain of genotype A/A ; C/C ; R/R ; pr/pr with a strain of genotypea/a ; c/c ; r/r ; Pr/Pr.a. What are the phenotypes of the parents?b. What will be the phenotype of the F1?c. What phenotypes, and in what proportions, willappear in the progeny of a selfed F1?d. What progeny proportions do you predict from thetestcross of an F1?6. Suppose that brown is the wild-type colour of the iris of eyes. However, dominant autosomal mutations in the genes coding the proteins responsible for the steps of colour determination (shown below) negatively disrupts their functions. The colour of the iris can be either blue, green, brown or no coloration (results in pink iris). B gene represents the pigment transporter; L gene represents the luciferase enzyme; and the C gene represents tyrosinase enzyme). The biochemical pathway for colour determination is the following: Cell of the human iris Blue Blue Luciferase (L) Green Pigment Tyrosinase (C) transporter (B) Brown a. Assuming no confounding factors, which genotype (include all genes mentioned) will result in green iris? b. Assuming that the transporter is composed of 4 subunits of protein coded by the same gene. What will the colour of the iris be if an individual is heterozygote for the gene expressing the transporter? c. Now, assuming the tyrosinase step is incompletely…
- 3. In 1988, neurologists in Australia reported the existence of identical twins who had developed myoclonic epilepsy in their teens. One twin remainedonly mildly affected by this condition, but the othertwin later developed other symptoms of full-blownMERRF, including deafness, ragged red fibers, andataxia (loss of the ability to control muscles).Explain the phenotypic dissimilarity in theseidentical twins.A rare dominant mutation expressed at birth was studiedin humans. Records showed that six cases were discoveredin 40,000 live births. Family histories revealed that in twocases, the mutation was already present in one of the parents.Calculate the spontaneous mutation rate for this mutation.What are some underlying assumptions that may affect ourconclusions?4. The A and 1B alleles in human blood are co-dominant but both are completely dominant to the i (give O blood type) allele. Fill in the table: Phenotype Genotype(s) A blood OR B blood OR AB blood poojq o Problem: Can a man who has O blood father a child with B blood? Explain your answer. Personet use only, do mOeproduce cd.eou garciato187 5 Hemophilia is a sex-linked recessive gene found on the X chromosome. A woman who is heterozygous for the gene has children with a man who has the condition. a What is the genotype of the man? What is the genotype of the woman? b Gametes produced by the man? Gametes produced by the woman? C In the space provided, draw a Punnett square for the parents previously described in this question. Include phenotype in each box. Next to your Punnett square, clearly identify the ratios for both genotype and phenotype. Personalise ony cootreoroduIce 70 TT-0207 GAICiato187@student.laccd.edu Personal use oniv do not reproduce 20-TT-0707 garciafo187@student…