Test 2, Deck 8

Question 1

Erminette chickens are fleckered white and black. A cross of two erminettes made 22 erminettes, 14 black and 12 whites, What is the genotype of erminette?

Answer 1

EE’

Question 2

What would be the genotype of an individual heterozygote with green eyes?

Answer 2

Gg

Question 3

What is the expected phenotype of an individual with the genotype gg for the green-eye gene?

Answer 3

blue eyes

Question 4

What would be the genotype of an individual homozygous recessive for the green eye gene?

Answer 4

gg

Question 5

Where is the probable locus of eye color alleles on chr of a genotype Gg ?

Answer 5

Each have to be on homologues pairs.

Question 6

If a brown eyed woman and a blue eye man produce 4 brown eyed children, what is the genotype of the man?

Answer 6

bb

Question 7

If a brown eyed woman and a blue eye man produce 4 brown eyed children, what is the genotype of the children ?

Answer 7

Bb

Question 8

If a brown eyed woman and a blue eye man produce 4 brown eyed children, what is the genotype of the woman?

Answer 8

B- (no info on second allele)

Question 9

If a woman BB and a bb man produces 4 children, what is the probability that all 4 children will have brown eyes?

Answer 9

100%

Question 10

If a Bb woman and a bb man reproduce 4 children, what is the probability that all 4 children will have brown eyes ?

Answer 10

1/16 ( 1/2x1/2x1/2x1/2)

Question 11

A mother and father find the taste of a chemical very bitter. 2/3 of their children find the chemical tasteless. Which phenotype is dominant?

Answer 11

Tasting it bitter, the parents are heterozygous

Question 12

A man and a woman are heterozygous for a recessive gene (albinism-no pigment). If they have twins, what is the probability that both infants have the same phenotype as parents?

Answer 12

if twins identical: 3/4

if twins non-identical: 9/16

Question 13

Dominant allele B = wild type color, recessive allele=b black color. A test-cross of a wild type female gave 52 black, 58 wild offspring. If the wild type females were crossed with their F1 black siblings, what genotypic ratios would be expected in F2?

Answer 13

Bb:bb 1:1

Question 14

Dominant allele B = wild type color, recessive allele=b black color.A test-cross of a wild type female gave 52 black, 58 wild offspring. If the wild-type females are crossed with their black F1 male siblings, what phenotypic ratios are expected in F2?

Answer 14

1: 1 black, wild-type

1/2 Bb, 1/2 bb

Question 15

Assume that an organism is heterozygous for 3 independent genes. If it is self-fertilized, what proportions of the offspring will carry AaBbcc?

Answer 15

1/16

Question 16

Assume that an organism is heterozygous, for 3 independent genes. If it is self-fertilized, what proportion of the offspring will have the same phenotype?

Answer 16

27/64 (AaBbCc x AaBbCc —> AaBbCc)

Question 17

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.

Answer 17

The regulatory element not functioning are the Cap and Camp proteins. 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. The Cap activator is only present if the CAMP is present, which it is if the glucose amount is low. And then, the transcription can begin. If B-galactosidase is in hight amount in the cell when both lactose and glucose is present, this means that the CAP activator and the CAMP were mutated because it is present and starts transcription event thought there is glucose.

Question 18

Match the types of mutation with the definition
Missense
Silent
Non-sense
Frameshift

Answer 18

Changes an amino acid
No change to amino acid
Creates a stop codon
insertion and deletion that changes the reading frame

Question 19

Identify the mutation
normal 3’ TAC GCA GAG ATC 5’
mutated 3’ TAC GCA GAC ATC 5’

Answer 19

CUC: Leu
CUG: Leu
Silent

Question 20

Select the mutation with the fewest change in amino acids:
A Single base deletion in middle of intron
B Single base insertion near the end of exon

Answer 20

A Single base deletion in middle of intron

Introns are removed from pre-mRNA by Spliceosomes so no effect on amino acid sequence.

Question 21

Select the mutation with the fewest change in amino acids:
A Base pair substitution that removes an intron splice site
B Single base insertion near the end of exon

Answer 21

B Single base insertion near the end of exon
This makes a frameshift for the rest of the exon, whereas removing an intron splice site retains the introns in mRNA creating a longer aa chain because in humans, introns are bigger than exons.