In a wild-type fungus, protein E (encoded by the haplosufficient gene “E”) normally homodimerizes, and the E-E dimer catalyzes a biochemical reaction necessary for the production of a dark pigment. ED represents a mutant, dominant, negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype E+/ED, and why?
A) mutant (no pigment production), as E is
haplosufficient
B) mutant (no pigment production), as no E-E dimers will form in
the heterozygous
C) mutant (no pigment production), as the mutant allele
ED is dominant (correct answer.
why??
D) wild type (normal production of the dark pigment), as
ED is a negative allele
E) wild type (normal production of the dark pigment), as E
is haplosufficient
In a wild-type fungus, haplosufficient gene encoding protein E undergoes dimerization, that is able to produce a dark pigment. However, haplosufficient gene refers to situations where a single copy of a functional gene has the ability to maintain normal function, and remains undented even after the cells make defected proteins. Moreover, ED mutant that is dominant and negative allele,which leads to dominant negative mutation in a dimeric membrane receptor, resulting in a truncation of the cytoplasmic domain (a part of signal transduction mechanism). Primarily,phenotypic characters of a species depend upon the allele that is dominant, and ED is dominant and negative allele in this case. Therefore phenotype of a fungus cell (having genotype E+/ED) will be mutant type with no pigment production.
In a wild-type fungus, protein E (encoded by the haplosufficient gene “E”) normally homodimerizes, and the...
Help with both please!
10. In a wild-type diploid fungus, protein E (encoded by the haplosufficient gene “E”') normally homodimerizes, and the E-E dimer catalyzes a biochemical reaction necessary for the production of a dark pigment. E represents a mutant, dominant negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype ET/E”, and why? A) mutant (no pigment production), as E is haplosufficient B) mutant (no pigment production), as no E-E dimers will form...
In a wild-type carnivorous plant, protein X is encoded by a haplosufficient gene “X”. The X protein normally homodimerizes. The X-X dimer enzymatically catalyzes a biochemical reaction that is critical for the digestion of insects that it traps. XD represents a dominant negative allele, a mutation, of gene X. What would be the predicted phenotype of a plant with the genotype X+/XD ? Which one of these pieces of information allows you to reach that conclusion? a. The plant would...
12) A particular allele of a gene (including the wild-type allele) can be expressed in a dominant fashion for several reasons A) Explain why wild-type alleles are usually dominant to mutant alleles (5 pts.). 20 B) Describe three ways in which mutant alleles can be dominant to a wild-type allele (6 pts.).
With respect to genetics, ABO blood types are controlled by a single gene designated I, which has three alleles. The I A allele gives blood type A, I B gives blood type B, and i gives blood type O. The I A and I B alleles are codominant and i allele is recessive to both the I A and I B alleles. Aside from the I alleles, there are two other genes that when mutated affect ABO blood type inheritance....
12) A particular allele of a gene (including the wild-type allele) can be expressed in a dominant fashion for several reasons- A) Explain why wild-type alleles are usually dominant to mutant alleles (5 pts.). B) Describe three ways in which mutant alleles can be dominant to a wild-type allele (6 pts.). 13) Explain how chromatids/chromosomes move to opposite poles of the cell during anaphase. Be sure to mention in your answer how the force that propels this movement is generated...
In zebrafish, the half baked (hab) gene is caused by a mutation in E-cadherin. A new allele of this gene, k51, has a wild-type phenotype as a heterozygote, and a strong lethal phenotype as a homozygote. However, k51/deletion has a slightly more dramatic phenotype than k51/k51. This test tells us that the new allele is: O a null allele O codominant. O dominant lethal. O temperature sensitive. O haploinsufficient. O a leaky allele O incompletely dominant. O dominant negative. O...
INTERACTIVE QUESTION 14.6 Consider an example in which the Mim gene (for pig- ment production) is epistatic to the B/b gene (for deposi- tion of pigment). Suppose that a dominant allele M is necessary for the production of the black pigment mela- nin; mm individuals are white. A dominant allele B re- sults in the deposition of a lot of pigment in an animal's hair, producing a black color. The genotype bb produces Phenotype Genotype Ratio Black M_bb I SS...
Copy 2 of In tigers, 24 units of enzyme activity are needed for the wild-type (striped) phenotype. Less than 24 units of activity results in a solid colored tiger. Al is the WT allele and produces an enzyme with 16 activity units A2 is the mutant allele and produces an enzyme with 10 activity units Which allele is dominant, and what is the phenotype of a heterozygous tiger? Explain your answer and state whether this is an example of haplosu...
In tigers, 24 units of enzyme activity are needed for the wild-type (striped) phenotype. Less than 24 units of activity results in a solid-colored tiger. A1 is the WT allele and produces an enzyme with 16 activity units A2 is the mutant allele and produces an enzyme with 10 activity units Which allele is dominant, and what is the phenotype of a heterozygous tiger? Explain your answer and state whether this is an example of haplosufficiency or haploinsufficiency (referring to...
In Drosophila (fruit flies) the genes how, dumpy and ebony are located on chromosome 3. LOF = loss of function. Flies homozygous for a LOF mutation (no gene product made) in ebony have dark black bodies. Flies homozygous for a LOF mutation (no gene product made) in dumpyhave truncated (short) wings. Flies homozygous for a partial LOF mutation (some gene product made but significantly less than normal) in how have wings that will not fold down (held out wings; that's...