Transcript Review-Questions-for-mendeliangenetics

Review Questions for
Mendelian Genetics
Match the description with the
genetics terms. Each question can
have more than one answer.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
The mixture of alleles in an individual A
A. genotype
The specific form of a gene. C
B. phenotype
For example, tall, short, white, purple B, C C. allele
Where a gene is located on a chromosome F D. gene
For example, BB or Tt A
E. Chromosome
Homozygous dominant A
F. Locus
heterozygous A
hybrid A
For example, B or b C
For example, flower color D
For example, green seeds or yellow seeds B
Can contain multiple genes E
The specific genetic makeup of an individual A
Match the description with the
genetics terms. Each question can
have more than one answer.
1. Genotype of an individual with a dominant
phenotype
D, E
2. Requires mating with a recessive individual A
3. Has the alleles G and t from an individual
A. Test cross
that was GgTt B
B. Gamete
4. Used to determine the possible offspring
from a cross C
C. Punnett Square
5. Genotype of an individual with the F
D. Heterozygous
recessive trait
E. Homozygous
6. A way to determine the genotype of an
dominant
individual with the dominant trait A
F. Homozygous
7. Contains two different alleles D
recessive
8. For example, hh F
9. hybrid D
7 Steps to doing genetics problems
•
•
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•
Define alleles
Parental genotypes
Gametes that parents can produce
Punnett Square using gametes
Fill in Punnett Square to determine offspring
genotypes
• Write genotype ratios & phenotype ratios of
offspring
• Answer the question and circle your answer.
If parents are Aabb and AaBb, assuming
independent assortment and random
recombination, what is the chance that
they
have
a
child
who
is
aaBb?
• Determine the gametes that each
parent can produce.
Ab
ab
AB
Ab
aB
ab
• Create your Punnett Square.
AABb
AAbb
AaBb
Aabb
AaBb
Aabb
aaBb
aabb
• Answer the question
1/8 or 12.5% chance of having aaBb child
If parents are aaBB and AaBb, assuming
independent assortment and random
recombination, what is the chance that
they have a child who is AaBb?
Use the product law:
• Chance of aa X Aa making Aa:
Possible
gametes:
a
A
a
• Chance of BB X Bb making Bb:
Possible
gametes:
B
B
b
Aa ½ chance
aa
BB
of getting
Aa
½ chance
of getting
Bb
• Multiply the two probabilities Bb
to get your answer:
(1/2)(1/2) = ¼ or 25% of having AaBb child
If parents are AaBB and AaBb, assuming
independent assortment and random
recombination, what is the chance that
they have a child who is aaBb?
Use the product law:
• Chance of Aa X Aa making aa: AA Aa
A
A
a
a
Aa
aa
• Chance of BB X Bb making Bb:
B
B
b
BB
1/4 chance
of getting
aa
½ chance
of getting
Bb
• Multiply the two probabilities to Bb
get your answer:
(1/4)(1/2) = 1/8 or 12.5% chance of having aaBb child
If parents are Aabb and AaBb, assuming
independent assortment and random
recombination, what is the chance that
they have a child who is AaBb?
Use the product law:
• Chance of Aa X Aa making Aa: AA Aa
A
A
a
a
Aa
aa
• Chance of BB X Bb making Bb:
b
B
b
Bb
1/2 chance
of getting
Aa
½ chance
of getting
Bb
• Multiply the two probabilities to bb
get your answer:
(1/2)(1/2) = 1/4 or 25% chance of having AaBb child
In peas, yellow color pea is dominant to green. Tall is
dominant to short. What percent of the offspring of two
individuals heterozygous for both traits is expected to be
tall and green?
Define alleles: Yellow = A; green = a
tall = T; short = t
Parental genotypes: Both AaTt
Use the product law:
• Chance of Aa X Aa making green:AA Aa
A
a
A
a
Aa
• Chance of Tt X Tt making tall:
T
t
T
t
TT
• Multiply the two probabilities to Tt
get your answer:
aa
1/4 chance
of getting
aa that is
green
3/4 chance
Tt of getting
tt tall (both
TT and Tt)
(1/4)(3/4) = 3/16 chance of having tall green pea
If parents are AabbDd and AaBbDd, assuming
independent assortment and random
recombination, what is the chance that they
have a child who is aaBbDD?
Use the product law:
• Chance of Aa X Aa making aa:
A
a
A
a
• Chance of bb X Bb making Bb:
b
•
•
B
b
AA Aa 1/4 chance
Aa
aa
of getting
aa
Bb ½ chance
of getting
Chance of Dd X Dd making DD:
bb Bb
D
d
d
D
DD Dd 1/4 chance
of getting
Multiply the 3 probabilities to get
Dd dd DD
your answer:
(1/4)(1/2)(1/4) = 1/32 chance of having aaBbDD child
If one parent has the genotype AabbDd
assuming independent assortment and
random recombination, which of the
following are possible gametes made by that
a. Aa
individual?
b.
c.
d.
e.
f.
g.
h.
AbD
ABd
Aabbdd
D
d
B
abd
What is the chance that
that parent would produce
the gamete AbD?
½ for A, 1 for b, ½ for D, so all
together ¼ chance to
produce AbD