Homework Answers
Answer:- A. 3 wild type and 4 poky, ad7-
Explanation:- By crossing the ad7- with the wild strain we get 3 wild type and 4 poky means weak.
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In a haploid fungus, Neurospora, the poky mutation (weak growth) is inherited mitochondrially, while ad7 is...
mutant (petite #3), the nuclear mutation in the segregational petite mutant will segregate in a 22 ratio. Since ne...
mutant (petite #3), the nuclear mutation in the segregational petite mutant will segregate in a 22 ratio. Since neutral petites lack mitochondrial DNA, all of the mitochondria in the offspring will come from the segregational petite parent. Thus, half of the offspring will be wild type and half will be mutant Part C Several isolated haploid petite mutants and wild-type cells were mated with each other on complete media to form diploid cells. The diploid cells were then starved so...
14..........In Neurospora (a fungus), a complementation test was performed on tive haploid mutants Wild-type in appearance,...
14..........In Neurospora (a fungus), a complementation test was performed on tive haploid mutants Wild-type in appearance, and means showed abnormal branching. (3 pts) It the table below, complementation crosses were made to establish which mutants rescue the mutation. A complementation group will contain mutants that can rescue the defective phenotype. OWN These results indicate that the mutations were in A) 1 gene. B) 2 genes. C) 3 genes. D) 4 genes. The polarity of DNA synthesis is: E) 5 genes....
need help with question 2, 4, 6 a.&b. 2. For a (fictitious) haploid fungus, the starting...
need help with question 2, 4, 6 a.&b.
2. For a (fictitious) haploid fungus, the starting point in the biosynthesis of the amino acid arginine is Compound X, which is always present in, and absorbed from the environment. The arginine biosynthetic pathway is: Enzyme B → Compound Y Compound z Enzyme C Compound X Enzyme A arginine A mutant strain of genotype a (lacking enzyme A) is crossed to a mutant strain of genotype c (lacking enzyme C). What proportion...
Answers are C,A. Please explain the method to fix this kind of questions. Thanks 18. Shown...
Answers are C,A. Please explain
the method to fix this kind of questions. Thanks
18. Shown below are the results of a series of coinfections of E. coli B using T4 rll- strains similar to those employed by Benzer in his deletion mapping experiment. Each strain contains a different deletion mutation. The ability to produce wild-type progeny phage is indicated by (+), (O) indicates no wild-type progeny. + + OC Which order of deletions is most consistent with this data?...
As a student project, you have isolated six new mutant strains of E. coli with altered...
As a student project, you have isolated six new mutant strains
of E. coli with altered behavior of the lactose operon. The strains
are listed in the table below, together with their phenotypes (with
regard to significant ?-galactosidase synthesis) in three specific
situations.
Columns 1 and 2 present the phenotypes of each mutant haploid
strain. In column 1, the mutant is in an otherwise wild-type
genome. In column 2, the genome also carries a nonsense-suppressor
mutation (that is not present...
all please Question 2 (1 point) ✓ Saved In Drosophila, the mutant black (b) has a...
all please
Question 2 (1 point) ✓ Saved In Drosophila, the mutant black (b) has a black body and the wild-type (b+) has a gray body; the mutant vestigial (v) has wings that are short and crumpled compared the long wild-type wings (v+). These genes are linked and are located on the X- chromosome. A cross between a female fly and a black, vestigial winged male fly produced the following progeny: gray (b+), normal (v+) 20 gray (b+), vestigial (v)...
mutant (petite #3), the nuclear mutation in the segregational petite mutant will segregate in a 22 ratio. Since neutral petites lack mitochondrial DNA, all of the mitochondria in the offspring will come from the segregational petite parent. Thus, half of the offspring will be wild type and half will be mutant Part C Several isolated haploid petite mutants and wild-type cells were mated with each other on complete media to form diploid cells. The diploid cells were then starved so...
14..........In Neurospora (a fungus), a complementation test was performed on tive haploid mutants Wild-type in appearance, and means showed abnormal branching. (3 pts) It the table below, complementation crosses were made to establish which mutants rescue the mutation. A complementation group will contain mutants that can rescue the defective phenotype. OWN These results indicate that the mutations were in A) 1 gene. B) 2 genes. C) 3 genes. D) 4 genes. The polarity of DNA synthesis is: E) 5 genes....
need help with question 2, 4, 6 a.&b.
2. For a (fictitious) haploid fungus, the starting point in the biosynthesis of the amino acid arginine is Compound X, which is always present in, and absorbed from the environment. The arginine biosynthetic pathway is: Enzyme B → Compound Y Compound z Enzyme C Compound X Enzyme A arginine A mutant strain of genotype a (lacking enzyme A) is crossed to a mutant strain of genotype c (lacking enzyme C). What proportion...
Answers are C,A. Please explain
the method to fix this kind of questions. Thanks
18. Shown below are the results of a series of coinfections of E. coli B using T4 rll- strains similar to those employed by Benzer in his deletion mapping experiment. Each strain contains a different deletion mutation. The ability to produce wild-type progeny phage is indicated by (+), (O) indicates no wild-type progeny. + + OC Which order of deletions is most consistent with this data?...
As a student project, you have isolated six new mutant strains
of E. coli with altered behavior of the lactose operon. The strains
are listed in the table below, together with their phenotypes (with
regard to significant ?-galactosidase synthesis) in three specific
situations.
Columns 1 and 2 present the phenotypes of each mutant haploid
strain. In column 1, the mutant is in an otherwise wild-type
genome. In column 2, the genome also carries a nonsense-suppressor
mutation (that is not present...
all please
Question 2 (1 point) ✓ Saved In Drosophila, the mutant black (b) has a black body and the wild-type (b+) has a gray body; the mutant vestigial (v) has wings that are short and crumpled compared the long wild-type wings (v+). These genes are linked and are located on the X- chromosome. A cross between a female fly and a black, vestigial winged male fly produced the following progeny: gray (b+), normal (v+) 20 gray (b+), vestigial (v)...
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