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A man is a carrier for sickle cell anemia (Aa) and Tay-Sachs disease (Tt). Let’s say the genes involved in both of these disorders are on the same chromosome. Which of the following are possible representations of his genotype:
A
a
T
t
A
a
t
T
A. #1 only
B. #2 only
C. #1 and #2
#1 #2
Yellow pea color (Y) is dominant to green pea color (y). Round seed shape (R) is dominant to oval pea shape (r). Let’s say the genes responsible for both of these phenotypes are on the same chromosome. Which of the following genotypes should result in a yellow pea plant with round seeds?
A. YyRr
B.
C.
D. Both A and B
E. All of the above
Y
y
R
r
Y
y
r
R
A man is a carrier for sickle cell anemia (Aa) and Tay-Sachs disease (Tt). Let’s say the genes involved in both of these disorders are on the same chromosome. Which of the following are possible representations of his genotype:
#1 #2 #3
AaTt
A
a
T
t
A
a
t
T
A. #1 only
B. #1 and #2
C. #1 and #3
D. #2 and #3
E. #1, #2, and #3
155 Wild-type
165 muscle defects, shriveled wings
330 muscle defects, normal wings
350 normal muscles, shriveled wings
Wild-type female Muscle defects, shriveled wing male
tz = mutant version of tafazzin
tz+ = normal version of tafazzin
vg = mutant version of vestigial
vg+ = normal version of vestigial
Normal dominant over mutant for both
tz
tz
vg
vg
What is the genotype of the wild-type parental female fly?
A.
B.
C. There’s not enough information
tz
tz+
vg
vg+
tz
tz+
vg+
vg
X
155 Wild-type
165 pink eye, shriveled wings
330 pink eye, normal wings
350 normal eyes, shriveled wings
Wild-type female Pink eye, shriveled wing male
p = mutant version of pink
p+ = normal version of pink
wg = mutant version of wing
wg+ = normal version of wing
Normal dominant over mutant for both
p
p
wg
wg
What do the chromosomes of the wild-type parental female fly look like?
A.
B.
C. There’s not enough information
p
p+
wg
wg+
p
p+
wg+
wg
X
In the test cross above what is the arrangement of the alleles?
A b
a B
a b
a b
a b
A B
a b
a b
A B
a b
a b
a b D. A or B
E. A,B, or C
X
X
X
A.
A.
B
C.
Adrenoleukodystrophy (ALD)
Accumulation of long fatty acid chains in the body
Neurodegeneration, problems with vision, hearing, and motor coordination
In Drosophila bubblegum gene is similar to the human gene that when mutant causes ALD.
You want to study the linkage between bubblegum and another gene, curly
Wild-type femaleCurly wings, vision
defects male
395 Wild-type
405 curly wings, vision defects
99 Vision defects, normal wings
101 curly wings, normal vision
Which classes of progeny arose from parental type gametes?
A. Wild-type & Vision defects, normal wings
B. Wild-type & Curly wings, vision defects
C. Vision defects, normal wings & Curly wings, normal vision
D. Curly wings, vision defects & Curly wings, normal vision
bgm+
bgm
cu+
cu
bgm
bgm
cu
cuX
In humans: problems with the heart, susceptibility to infections, and muscle
weakening
Mutations in tafazzin cause muscle weakness in flies.
A Drosophila gene called tafazzin is similar to the human gene that when mutant causes Barth syndrome.
Mutations in the gene vestigial prevents flies from normally spreading out their wings.
tz = mutant version of tafazzin
tz+ = normal version of tafazzin
vg = mutant version of vestigial
vg+ = normal version of vestigial
Normal dominant over mutant for both
Let’s look at the linkage between tafazzin and vestigial genes
What is the recombination frequency between curly and purple?
A. 0.16
B. 0.32
C. 0.49
D. 0.52
E. 0.68
155 Wild-type
165 curly wings, purple eyes
330 normal wings, purple eyes
350 curly wings, normal eyes
Wild-type female
c+ c, pr+ pr
Curly wings, purple eyes
c c, pr pr
1,000 total progeny
c = mutant version of curly
c+ = normal version of curly
pr = mutant version of purple
pr+ = normal version of purple
Normal dominant over mutant for both
X
Progeny:
What is the recombination frequency between black and green?
A. 0.15
B. 0.13
C. 0.28
D. 0.51
E. 0.72
Wild-type female
b+ b, g+ g
Black, green eye male
b b, g g
100 total progeny
X
36 wild-type
Progeny:
36 black, green eyes
15 gray, green eyes
13 black, white eyes
b = mutant, black body
b+ = normal, gray body
g = mutant, green eye
g+ = normal, white eye
Normal dominant over mutant for both
What is the recombination frequency between black and green?
A. 0.15
B. 0.13
C. 0.28
D. 0.49
E. 0.72
Wild-type female
b+b g+g
Black, green eye male
bb gg
100 total progeny
X
15 wild-type
Progeny:
13 black, green eyes
36 gray, green eyes
36 black, white eyes
b = mutant, black body
b+ = normal, gray body
g = mutant, green eye
g+ = normal, white eye
Normal dominant over mutant for both
What is the recombination frequency between curly and orange
A. 0.16
B. 0.31
C. 0.49
D. 0.50
E. 0.69
150 Wild-type
Wild-type female
c+c o+o
Curly wings, orange eyes
c c o o
1,000 total progeny
c = mutant version of curly
c+ = normal version of curly
o = mutant version of orange
o+ = normal version of orange
Normal dominant over mutant for both
160 curly wings, orange eyes
340 normal wings, orange eyes
350 curly wings, normal eyes
X
Progeny:
758 y+ w+ m+700 y w m401 y+ w+ m317 y w m+16 y+ w m12 y w+ m+1 y+ w m+0 y w+ m
Let’s look at the y w m cross again
What are the NCO chromosomes?
A. w+ y+ m+ & w y m
B. w+ y+ m & w y m+
C. w+ y m+ & w y+ m
D. w+ y m & w y+ m+
758 y+ w+ m+700 y w m401 y+ w+ m317 y w m+16 y+ w m12 y w+ m+1 y+ w m+0 y w+ m
Let’s look at the y w m cross again
What is the order of the genes?
A. w - y - m
B. y - w - m
C. w - m - y
• Female flies phenotypically wild-type and heterozygous for each of three different mutations [curly wings (c), short bristles (b), and sepia eyes (s)] were crossed to male flies that have curly wings, are short bristles, and have sepia eyes. The number of progeny in eight different phenotypic classes is:
Which of the three genes (c,b,s) is in the middle?
A. Curly
B. Short Bristles
C. Sepia
Wild-type = 446 Sepia eyes & Curly wings= 10
Curly wings = 42 Sepia eyes & Short bristles = 40
Sepia eyes = 2 Short bristles & Curly wings = 1
Short bristles = 10 Sepia eyes, Curly Wings, & Short bristle = 449
• Female flies phenotypically wild-type and heterozygous for each of three different mutations [purple eyes (pr), dumpy wings (dp), and hairy (h)] were crossed to male flies that have purple eyes, have dumpy wings, and are hairy. The number of progeny in eight different phenotypic classes is:
Which of the three genes (c,b,s) is in the middle?A. hairy
B. purple
C. dumpy
Wild-type = 298 Dumpy wings & Hairy = 30
Hairy = 8 Vision defects & Purple eyes = 165
Purple eyes = 28 Dumpy wings & Purple eyes = 10
Dumpy wings = 161 Dumpy wings, Purple eyes, Hairy = 300
Before answering “Why?,” let’s do a linkage analysis (based on Fig. 5.9 in the book):
Female flies who appear wild-type are heterozygous ebony body (e) and short bristles (s) (Gray body (+) and long bristles (+) are wild-type and dominant) are crossed to ebony body, short bristle males. The progeny are 537 wild-type flies, 76 flies with wild-type bristles and ebony bodies, 75 flies with wild-type body color and short bristles and 542 flies with ebony body and short bristles.
What are the parental (NCO) chromosomes?
A. +,+ & +, sB. +,+ & e, +C. +,+ & e, sD. e,s & e, +E. e,s & +, sF. e, + & +, s
Before answering “Why?,” let’s do a linkage analysis (based on Fig. 5.9 in the book):
Female flies who appear wild-type are heterozygous ebony body (e) and short bristles (s) (Gray body (+) and long bristles (+) are wild-type and dominant) are crossed to ebony body, short bristle males. The progeny are 537 wild-type flies, 76 flies with wild-type bristles and ebony bodies, 75 flies with wild-type body color and short bristles and 542 flies with ebony body and short bristles.
What are the recombinant (SCO) chromosomes?
A. +,+ & +, sB. +,+ & e, +C. +,+ & e, sD. e,s & e, +E. e,s & +, sF. e, + & +, s
Before answering “Why?,” let’s do a linkage analysis (based on Fig. 5.9 in the book):
Female flies who appear wild-type are heterozygous ebony body (e) and short bristles (s) (Gray body (+) and long bristles (+) are wild-type and dominant) are crossed to ebony body, short bristle males. The progeny are 537 wild-type flies, 76 flies with wild-type bristles and ebony bodies, 75 flies with wild-type body color and short bristles and 542 flies with ebony body and short bristles.
What is the map distance between ebony and short?
A. 6.9 B. 10.9 C. 12.3
D. 14.0 E. 27.9
Let’s go back to the white, yellow, miniature cross and check the gene order. Here’s the data for crossing a phenotyically wild-type female by a white eyed (w), yellow bodied (y), miniature winged (m) male:
2278 w+ y+ m What are the NCO chromosomes?
2157 w y m+ A. w+ y+ m+ & w y m
1203 w y m B. w+ y+ m & w y m+
1092 w+ y+ m+ C. w+ y m+ & w y+ m
49 w+ y m D. w+ y m & w y+ m+
41 w y+ m+
2 w+ y m+
1 w y+ m
Let’s go back to the white, yellow, miniature cross and check the gene order. Here’s the data for crossing a phenotyically wild-type female by a white eyed (w), yellow bodied (y), miniature winged (m) male:
2278 w+ y+ m What are the DCO chromosomes?
2157 w y m+ A. w+ y+ m+ & w y m
1203 w y m B. w+ y+ m & w y m+
1092 w+ y+ m+ C. w+ y m+ & w y+ m
49 w+ y m D. w+ y m & w y+ m+
41 w y+ m+
2 w+ y m+
1 w y+ m
Let’s go back to the white, yellow, miniature cross and check the gene order. Here’s the data for crossing a phenotyically wild-type female by a white eyed (w), yellow bodied (y), miniature winged (m) male:
2278 w+ y+ m What is the order of the genes?
2157 w y m+ A. w - y - m
1203 w y m B. y - w - m
1092 w+ y+ m+ C. w - m - y
49 w+ y m
41 w y+ m+
2 w+ y m+
1 w y+ m
Let’s go back to the white, yellow, miniature cross and check the gene order. Here’s the data for crossing a phenotyically wild-type female by a white eyed (w), yellow bodied (y), miniature winged (m) male:
2278 w+ y+ m
2157 w y m+ What is the map distance for w to m?
1203w y m A. 1.4
1092 w+ y+ m+ B. 33.6
49 w+ y m C. 33.7
41 w y+ m+ D. 34.1
2 w+ y m+ E. 35.0
1 w y+ m
Wild-type female fly, heterozygous for +/a, +/b and +/c is crossed to an a, b, c male.
The progeny include:
NCOs: abc and +++
DCOs: ab+ and ++c
What is the map?:
A. a - b - c
B. a - c - b
C. b - a - c
Wild-type female fly, heterozygous for +/a, +/b and +/c is crossed to an a, b, c male.
The progeny include:
NCOs: a+c and +b+
DCOs: ++c and ab+
What is the map?:
A. a - b - c
B. a - c - b
C. b - a - c
A recessive mutation in the X-linked yellow gene of Drosophila confers a yellow body color to the flies, and a recessive mutation in another X-linked gene, singed, gives singed hairs.
Let’s say you are working with a fly that has the genotype: y sn+/ y+sn, which of the possible phenotypes could you see as a result of mitotic recombination if a single crossover occurs between the sn gene and the centromere?A. a spot that is both yellow and singed B. a yellow spotC. a singed spotD. a twin spot
s +
+ y
A smurf with a wild type phenotype is heterozygous for the red gene (r+r) and the hairy gene (h+h). His chromosomes are shown below:
r h+
r+ h
If mitotic crossing over occurs between the r and h genes, what phenotypes are possible?
A) a red spotB) a hairy spotC) a red and hairy spotD) a twin spot (a red spot next to a hairy spot)E) only wild type