The amount of enzyme activity in a cell that is homozygous for a mutant allele is 0 units. The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 0 units. What type of allele is the mutant allele?
a. Dominant negative
b. Hypomorphic
c. null/amorphic
d. neomorphic
e. hypermorphic
in genetics, null or amorphic alleles are the alleles which have the same phenotype when homozygous as when heterozygous. so in both situation, these alleles have the same phenotype with a deficiency that disrupts the locus. the amorphic allele can be RNA or protein null but some time can express a defective protein which is nonfunctional.
in the above question, the allele has the same 0 unit activity in both homozygous and heterozygous condition hence it is amorphic/null allele
The amount of enzyme activity in a cell that is homozygous for a mutant allele is...
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...
20. A corn plant is homozygous for a mutant allele that results
in no pigment in the seed (i.e.,
white). The mutant is caused by Ds insertion that often exits late
in seed development, when
there is an active Ac element in the genome. If there is NO active
Ac element, the seeds of
this plant will be:
A) no pigment (i.e., white)
B) pigmented all over
C) white with small spots of pigment
D) white with large spots of...
Homework 7 due Monday, March 16 by 11:59pm Chapter 4 Question 14 Multiple Choice 8 of 10 - Part A The amount of enzyme activity in a cell that is homorygous for a mutant unts. What type of all is the mutant ? s 400 units The amount of anyme activity in a collomorygous for the Wallis 200 units. The amount of entryme activity in a h rygis 300 typermorphic neomorphie hypomorphic O dominant negative nullamorphic Submit Prin s Besvest...
A mutant plant with white flowers exists that lacks red anthocyanin pigment, normally made by enzyme P. Indeed the petal tissue lacks all detectable activity for enzyme P. Despite the lack of enzyme activity, a study of homozygous mutant cells using antibodies against the wild-type enzyme demonstrated that the cells homozygous for the mutation still had the enzyme (i.e., the antibody showed the presence of the enzyme). Which statement below could most likely explain these results? A )The mutant had...
4. In Drosophila melanogaster, the phenotype curly wings is due to a mutant allele Cy that is lethal when homozygous. A population is established with an initial frequency of Cy equal to 0.168. Denoting + as the wild- type (i.e. non-mutant) allele at this locus, calculate the expected frequency of Cy in the next generation if the relative fitness of the ++ homozygote to the Cy/+ heterozygote is: a) 1 : 1 b) 1 : 0.5 c) Briefly explain why...
4. A man makes 0 units of an enzyme due to homozygous recessive gene on chromosome 21. He marries a woman who produces 100 units of the same enzyme due to the heterozygous pairs of genes on her chromosomes 21. Their firstborn son has three copies of chromosome 21 and produces 200 units of the enzyme mentioned above. a) In which parent did the nondisjunction occur? b) Did the nondisjunction occur in meiosis I, meiosis II or mitosis? c) What...
4. A man makes 0 units of an enzyme due to homozygous recessive gene on chromosome 21. He marries a woman who produces 100 units of the same enzyme due to the heterozygous pairs of genes on her chromosomes 21. Their firstborn son has three copies of chromosome 21 and produces 200 units of the enzyme mentioned above. a) In which parent did the nondisjunction occur? b) Did the nondisjunction occur in meiosis I, meiosis II or mitosis? c) What...
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...
Alcohol dehydrogenase is an enzyme composed of several polypeptide subunits. It is normally expressed at high levels in the liver and the lining of the stomach. It is essential to metabolize alcohol, as without it, drinkers would become drunk faster and stay drunk longer. A scientist is studying mutations in mice that affect the activity of the enzyme. She finds that a heterozygote having a normal allele and a mutation that deletes the gene has 50% of normal enzyme activity....