Download genetics answers

BIOL 153L General Biology II Lab
Black Hills State University
Genetics Problem Answers (Exam 1 Study Guide)
1. If a plant is heterozygous for stem size (L = long stem allele, l = short stem allele) and for fruit color
(W = white fruit allele, w = purple fruit allele), what is its genotype?
LlWw
2. If the plant above had the genotype LLww, what is its phenotype? How about LlWw?
LLww: Long, purple
LlWw: Long, white
3. Two plants have the genotype BbGg, where B = big flower allele, b = little flower allele; G = green
leaf allele, g = maroon leaf allele)
• Assuming that the flower size and leaf color genes are NOT linked together on a chromosome,
what types of gametes could each plant make?
BG, Bg, bG, bg
• Assuming flower size and leaf color genes ARE linked, that the alleles have a cis arrangement, and
that no crossover between the two genes occur—what types of gametes are produced?
BG, bg
• If instead the alleles have a trans arrangement and there is no crossover between the two linked
genes, what types of gametes are produced?
Bg, bG
• If the alleles have a cis arrangement and there are crossovers between sister chromatids (only) in a
cell undergoing meiosis, what types of gametes are produced?
BG, bg
• If the alleles have a cis arrangement and there is one (and only one) crossover between linked
genes on adjacent non-sister chromatids in a cell undergoing meiosis, what types of gametes are
produced?
BG, Bg, bG, bg
1
4. Make a Punnett Square for the monohybrid cross involving parents with the genotype Bb. What is
the expected phenotypic and genotypic ratio for this cross?
(from mom) B
(from dad) B
BB
(from dad) b
Bb
(from mom) b
Bb
bb
Phenotypic ration = 3: 1 (3 big flower : 1 little flower)
Genotypic ratio = 1:2:1 (1 BB : 2 Bb : 1 bb)
5. Make a Punnett Square for a dihybrid cross involving parents with the genotype BbGg. What is the
expected phenotypic ratio for this cross?
BG
Bg
bG
bg
BG
BBGG
BBGg
BbGG
BbGg
Bg
BBGg
BBgg
BbGg
Bbgg
bG
BbGG
BbGg
bbGG
bbGg
bg
BbGg
Bbgg
bbGg
bbgg
9:3:3:1 (9 big green : 3 big maroon : 3 little green : 1 little maroon)
6. Make a Punnett Square for the linked gene scenarios listed under #3 above. What are the expected
phenotypic ratios for these crosses?
cis linkage
BG
bg
BG
bg
BBGG
BbGg
BbGg
bbgg
predicted cis- linkage ratio = 3:0:0:1 (3 big green, 0 big maroon, 0 little green, 1 little maroon)
trans linkage
Bg
bG
Bg
bG
BBgg
BbGg
BbGg
bbGG
predicted trans- linkage ratio = 2:1:1:0 (2 big green, 1 big maroon, 1 little green, 0 little maroon)
7. Consider the aforementioned crosses of plants with the genotype BbGg (see #2 above). Given no
prior knowledge about the location of the flower size and leaf color genes on the chromosome, use the
following possible results from the dihybrid cross to predict whether the two genes are linked (and if so,
whether the alleles are associated in cis or trans arrangements). (Circle the best answers.)
Case 1: 108 (big, green) : 36 (big, maroon) : 36 (little, green) : 12 (little, maroon)
[linked] or [unlinked]
if linked: [cis] or [trans]
and [recombination] or [no recombination]
Simplified ratio = 9:3:3:1
Matches predicted ratio for dihybrid cross with unlinked genes
2
Case 2: 144 (big, green) : 0 (big, maroon) : 0 (little, green) : 48 (little, maroon)
[linked] or [unlinked]
if linked: [cis] or [trans]
and [recombination] or [no recombination]
Simplified ratio = 12:0:0:4 (= 3:0:0:1)
Does not match 9:3:3:1 ratio so probably linked
Matches predicted ratio for cross with cis linkage (if you forget ratio you can make a Punnet
Square for cis and trans to remind yourself of expected ratios)
“Middle phenotypes” are zero so no recombination
Cross 3: 96 (big, green) : 48 (big, maroon) : 48 (little, green) : 0 (little, maroon)
[linked] or [unlinked]
if linked: [cis] or [trans] and [recombination] or [no recombination]
Simplified ratio = 8:4:4:0 (=2:1:1:0)
Does not match 9:3:3:1 ratio so probably linked
Matches predicted ratio for cross with trans linkage (if you forget ratio you can make a Punnet
Square for cis and trans to remind yourself of expected ratios)
“Doubly recessive” phenotype is zero so no recombination
Cross 4: 139 (big, green) : 6 (big, maroon) : 4 (little, green) : 43 (little, maroon)
[linked] or [unlinked]
if linked: [cis] or [trans]
and [recombination] or [no recombination]
Simplified ratio = 11.6 : 0.5 : 0.3 : 3.6
Does not match 9:3:3:1 ratio so probably linked
Rarest phenotypes are in “the middle” so most likely cis linkage
Rarest phenotypes are not zero so recombination occurred
8. Transcribe the following antisense strand of DNA into mRNA: TAC GAG ATA CGA TAC
AUG CUC UAU GCU AUG
9. What tRNA anticodons would attach to the aforementioned mRNA strand?
UAC GAG AUA CGA UAC
3