Test 2, Deck 6 Flashcards
Four different crosses of pea plants produces :
Cross Red peas blue peas
1 152 0
2 0 210
3 24 76
4 76 80
Select all the possible genotypes of the parents of cross 3?
Bb x Bb
Consider the cross of a homozygous red mouse with a heterozygous white mouse. Which phenotype is dominant?
White
If the two heterozygous mice are mated, what proportion of the offspring would be expected to have the same phenotype as the parent of a gene with complete dominance?
3/4
Provide the genotype of a short-hair black male cat with white paws.
Short-hair: L- (LL or Ll) Black make: XoY Spotting: S- (SS or Ss) Completely white fur: ww Dark pigment (black): D- (DD or Dd) Solid color: Aa (pattern)
Identify the type of mutation:
normal 3’ TAC GCC TCC ATC AGG 5’
mutated 3’ TAC GCC TCA TCA GG 5’
Frameshift mutation
Identify the type of mutation.
normal 3’ TAC AAC AGA ATC AGG 5’
mutated 3’ TAC AAG AGA ATC AGG 5’
Missence mutation
Rank the mutations in order from the those that cause the most changes to those that cause the fewest changes to the amino acid sequence.
Asingle base deletion in the middle of an intron
Bbase pair substitution that removes an intron splice site
Csingle base insertion near the end of an exon
Dbase pair substitution in an exon
most B[base pair substitution that removes an intron splice site]: next C[single base insertion near the end of an exon]: next D[base pair substitution in an exon]: fewest A[single base deletion in the middle of an intron]
point mutations within an intron would not be retained in the mRNA
a base substitution changes only a single codon
an insertion near the end of an exon changes all of the codons downstream from the insertion
removing an intron splice site would retain the intron in the mRNA creating a larger mRNA. Since in mammals, introns are often much larger than exons, this mutation would would change the greatest number of amino acids
The following diagram represents the structure of a gene in Drosophila melanogaster. Which segments of the gene will be represented in the primary transcript? Select all that apply.
enhancer (a), promoter (b), intron (c), exon (d), intron (e), exon (f), intron (g), nothing (h), enhancer (I) and nothing (j)
the primary transcript includes exons and introns. the promoter is not transcribed
The correct answers are: C, D, E, F, G
The following diagram represents the structure of a gene in Drosophila melanogaster. Which segments of the gene will be removed by RNA splicing? Select all that apply.
enhancer (a), promoter (b), intron (c), exon (d), intron (e), exon (f), intron (g), nothing (h), enhancer (I) and nothing (j)
introns are removed from the primary transcript
E and G
The following diagram represents the structure of a gene in Drosophila melanogaster. Which segments would most likely bind proteins that interact with RNA polymerase? Select all that apply.
enhancer (a), nothing (b), promoter (c), exon (d), intron (e), exon (f), intron (g), nothing (h), enhancer (I) and nothing (j)
proteins can bind to enhancers and the promoter
The correct answers are: A, C, J
No ß-galactosidase was observed in a test tube containing E. coli growing in an environment with glucose but no lactose. Which of the following genetic elements could be mutated and no longer functional? Select all that apply.
a. this is normal, no mutations can be detected
b. lac Y
c. none of these genetic mutations could explain this observation
d. operator
e. crp (codes for CAP)
f. lac Z
g. lac I
A this is normal, no mutations can be detected
A large amount of ß-galactosidase was observed in a test tube containing E. coli growing in an environment with lactose and glucose. Which regulatory element(s) is(are) not functioning as expected? Briefly justify your choice.
Since lactose is present, the functionality of the repressor and operator cannot be determined. lactose will remove the repressor from the operator
In a normale cell, If both lactose and glucose are present, the lactose will be ignored whereas the glucose will be used because it is more efficient to break down glucose than lactose so if glucose is present. If B-galactosidase is present even if there is glucose and lactose, this means that the lactose was used and not the glucose. The CAP and CAMP were mutated because the CAP is present and starts translation even thought CAMP is only present when glucose is low
Select the eukaryotic gene regulatory mechanism(s) that can increase gene expression.
a. multiple copies of genes
b. alternative splicing
c. adding methyl groups near the promoter
d. enhancers
e. activators
f. RNAi
g. repressors
h. heterochromatin
a. multiple copies of genes
d. enhancers
e. activators
A child with blood type A has a mother with blood type O. What blood types could the father have had?
A or AB (Mom is IoIo (type O alleles), she can only donate O, father must have at least one A
Is it possible for two parents with type A blood to produce a child with O blood?
Yes, if both parents are heterozygous for the A allele (AO)
