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WorksheetsChapter 11 - "Monohybrid vs Dihybrid Genetic Cross"
Total questions: 23
Worksheet time: 2hrs 41mins
Name
Class
Date
1.
How many alleles (letters) do each person carry for a trait?
a)
1
b)
2
c)
3
d)
4
2.
How many alleles (letters) does a child get from each parent for each trait?
a)
1
b)
2
c)
3
d)
4
3.
Dominant alleles are represented by a:
a)
Male gene
b)
lowercase letter
c)
recessive trait
d)
capital letter
4.
Aa, DD, bB, yy are all examples of
a)
genotypes
b)
phenotypes
5.
T - tall and t = short
In the punnett square below, what belongs in the missing square
In the punnett square below, what belongs in the missing square
a)
tt
b)
Tt
c)
Bb
d)
TT
6.
AA and Aa always show up as
a)
dominant
b)
recessive
7.
aa always shows up as
a)
dominant
b)
recessive
8.
Aa is
a)
homozygous
b)
heterozygous
9.
Which of the following is a phenotype?
a)
Aa
b)
red hair
c)
heterozygous
d)
DNA
10.
Which of the following genotypes is homozygous dominant?
a)
A
b)
Aa
c)
AA
d)
aa
11.
Which of the following genotypes is heterozygous?
a)
Bb
b)
BB
c)
bb
d)
B
12.
Which of the following is homozygous recessive?
a)
Tt
b)
tt
c)
TT
d)
T
13.
What are the possible genotypes for this cross?
a)
TT = 25% Tt= 50% tt=25%
b)
all tall
c)
TT= 75% tt = 25%
d)
Tt = 100%
14.
Gregor Mendel, the botanist scientist, was breeding pea plants to find out the genotypes and phenotypes of future offspring. He decided to breed two pea plants who's flower genetic makeup were both heterozygous dominant (Tt). The dominant allele was considered blue, while the recessive allele was considered to be green. Create a punnet square chart by crossing the two heterozygous dominant traits, and find out the percent chance that the F1 would rather be blue flowers, or green flowers?
a)
The ratio of the F1 flower offspring being the color blue would be 2/4, and the ratio of the green flowers would be 2/4.
b)
The ratio of the F1 flower offspring being the color blue would be 3/4, and the ratio of the green flowers would be 1/4.
c)
The ratio of the F1 flower offspring being the color blue would be 1/4, and the ratio of the green flowers would be 3/4.
d)
All of the flowers would be blue in color, there would be no green flowers in the F1 offspring.
15.
Gregor Mendel, the botanist scientist, was breeding pea plants to find out the genotypes and phenotypes of future offspring. He decided to breed two pea plants who's flower genetic makeup were both heterozygous dominant (Tt). The dominant allele was considered blue, while the recessive allele was considered to be green. Create a punnet square by crossing the two heterozygous dominant traits, and find out the F1 genotype offspring that the plants would create?
a)
The F1 genotype offspring would consist of the following allele pairs: TT, TT, Tt, Tt
b)
The F1 genotype offspring would consist of the following allele pairs: TT, tt, tt, Tt
c)
The F1 genotype offspring would consist of the following allele pairs: TT, Tt, Tt, tt
d)
tt, tt, TT, TT
16.
JURASSIC WORLD WAS A NEW ATTRACTION BUILD IN 2011, WITHIN WHICH WAS A HUGE SUCESS AT FIRST. PEOPLE AROUND THE WORLD WOULD COME SEE MANY DINOSAURS. ONE OF THE DINOSAURS PEOPLE WOULD LIKE TO SEE IS THE FIRST "HUMAN-MADE GENETIC HYBRID" THE INDOMINOUS REX. PROFESSOR LLOYD WAS TRYING TO IDENTIFY THE GENOTYPES OF THE PARENT OFFSPRING FOR TEETH TYPE. WHAT WOULD THE GENOTYPES OF TWO PARENT INDOMINOUS REX BE IF ONE HAD A GENOTYPE OF HETEROZYGOUS DOMINANT FOR TEETH TYPE, AND THE OTHER GENOTYPE WAS HOMOZYGOUS RECESSIVE FOR TEETH TYPE?
a)
THE FIRST INDOMINOUS REX THAT HAD THE GENETIC MAKEUP OF THE HETEROZYGOUS DOMINANT TRAIT WOULD BE (Ee). ON THE OTHER HAND, THE GENETIC MAKEUP OF THE OTHER INDOMINOUS REX WOULD BE HOMOZYGOUS RECESSIVE (ee).
b)
THE FIRST INDOMINOUS REX THAT HAD THE GENETIC MAKEUP OF THE HETEROZYGOUS DOMINANT TRAIT WOULD BE (Ee). ON THE OTHER HAND, THE GENETIC MAKEUP OF THE OTHER INDOMINOUS REX WOULD BE HOMOZYGOUS RECESSIVE (EE).
c)
THE FIRST INDOMINOUS REX THAT HAD THE GENETIC MAKEUP OF THE HETEROZYGOUS DOMINANT TRAIT WOULD BE (EE). ON THE OTHER HAND, THE GENETIC MAKEUP OF THE OTHER INDOMINOUS REX WOULD BE HOMOZYGOUS RECESSIVE (ee).
d)
THE FIRST INDOMINOUS REX THAT HAD THE GENETIC MAKEUP OF THE HETEROZYGOUS DOMINANT TRAIT WOULD BE (ee). ON THE OTHER HAND, THE GENETIC MAKEUP OF THE OTHER INDOMINOUS REX WOULD BE HOMOZYGOUS RECESSIVE (ee).
17.
PROFESSOR LLOYD CONTINUED TO LOOK AT THE P1 GENERATION, AND WANTED TO IDENTIFY THE PHENOTYPES OF EACH PARENT FOR TEETH TYPE. WHAT WOULD BE THE PHENOTYPES OF EACH PARENT INDOMINOUS REX IF THE MALE PARENT DINOSAUR WAS HETEROZYGOUS DOMINANT FOR GENOTYPE, WHILE THE FEMALE INDOMINOUS REX PARENT WAS HOMOZYGOUS RECESSIVE FOR GENOTYPE?
a)
THE PHENOTYPE OF THE INDOMINOUS REX WITH THE GENOTYPE OF HETEROZYGOUS DOMINANT WOULD BE SHARK TEETH. ON THE OTHER HAND, THE PHENOTYPE OF THE OTHER INDOMINOUS REX WITH THE GENOTYPE OF HOMOZYGOUS RECESSIVE WOULD BE SHARK TEETH.
b)
THE PHENOTYPE OF THE INDOMINOUS REX WITH THE GENOTYPE OF HETEROZYGOUS DOMINANT WOULD BE ELEPHANT TUSKS. ON THE OTHER HAND, THE PHENOTYPE OF THE OTHER INDOMINOUS REX WITH THE GENOTYPE OF HOMOZYGOUS RECESSIVE WOULD BE SHARK TEETH.
c)
THE PHENOTYPE OF THE INDOMINOUS REX WITH THE GENOTYPE OF HETEROZYGOUS DOMINANT WOULD BE SHARK TEETH. ON THE OTHER HAND, THE PHENOTYPE OF THE OTHER INDOMINOUS REX WITH THE GENOTYPE OF HOMOZYGOUS RECESSIVE WOULD BE ELEPHANT TUSKS.
d)
THE PHENOTYPE OF THE INDOMINOUS REX WITH THE GENOTYPE OF HETEROZYGOUS DOMINANT WOULD BE ELEPHANT TUSKS. ON THE OTHER HAND, THE PHENOTYPE OF THE OTHER INDOMINOUS REX WITH THE GENOTYPE OF HOMOZYGOUS RECESSIVE WOULD BE ELEPHANT TUSKS.
18.
MR. LLOYD DECIDED TO BREED THE P1 GENERATION OFFSPRING OF INDOMINOUS REX. THE MALE INDOMINOUS REX HAD A GENOTYPE OF HETEROZYGOUS DOMINANT FOR TEETH TYPE, WHILE THE FEMALE INDOMINOUS REX HAD A GENOTYPE OF HOMOZYGOUS RECESSIVE FOR TEETH TYPE. WHAT WOULD BE THE RESULT THE F1 OFFSPRING'S TEETH GENOTYPE RATIO? HINT, YOU HAVE TO CREATE A PUNNET SQUARE CHART, OR FOIL METHOD TO FIND THE F1 OFFSPRINGS TEETH GENOTYPE PERCENTAGE.
a)
THE PERCENT CHANCE THAT THE F1 OFFSPRING WOULD HAVE THE GENOTYPE OF HETEROZYGOUS DOMINANT WOULD BE 25 PERCENT. ON THE OTHER HAND, THE PERCENT CHANCE THAT THE F1 OFFSPRING WOULD HAVE THE GENOTYPE OF HOMOZYGOUS RECESSIVE WOULD BE 50 PERCENT. THE FINAL DINO WOULD HAVE A 25 PERCENT CHANCE TO HAVE THE HOMOZYGOUS DOMINANT GENOTYPE.
b)
THE PERCENT CHANCE THAT THE F1 OFFSPRING WOULD HAVE THE GENOTYPE OF HETEROZYGOUS DOMINANT WOULD BE 50 PERCENT. ON THE OTHER HAND, THE PERCENT CHANCE THAT THE F1 OFFSPRING WOULD HAVE THE GENOTYPE OF HOMOZYGOUS RECESSIVE WOULD BE 25 PERCENT. THE FINAL DINO WOULD HAVE A 25 PERCENT CHANCE TO HAVE THE HOMOZYGOUS DOMINANT GENOTYPE.
c)
THE PERCENT CHANCE THAT THE F1 OFFSPRING WOULD HAVE THE GENOTYPE OF HETEROZYGOUS DOMINANT WOULD BE 25 PERCENT. ON THE OTHER HAND, THE PERCENT CHANCE THAT THE F1 OFFSPRING WOULD HAVE THE GENOTYPE OF HOMOZYGOUS RECESSIVE WOULD BE 75 PERCENT.
d)
THE PERCENT CHANCE THAT THE F1 OFFSPRING WOULD HAVE THE GENOTYPE OF HETEROZYGOUS DOMINANT WOULD BE 50 PERCENT. ON THE OTHER HAND, THE PERCENT CHANCE THAT THE F1 OFFSPRING WOULD HAVE THE GENOTYPE OF HOMOZYGOUS RECESSIVE WOULD BE 50 PERCENT.
19.
What genotype is missing from this Punnett Square?
a)
RrYy
b)
RRYY
c)
rryy
d)
RrYY
20.
Based off this punnett square, what fraction of the offspring will have wrinkled, yellow seeds?
a)
9/16
b)
3/16
c)
1/16
d)
16/16
21.
What is the genotype for a pea plant heterozygous for round seeds (R), and homozygous recessive for green seeds (y)?
a)
Ry
b)
RRyy
c)
RrYy
d)
Rryy
22.
Cross Two Homozygous plants
(RRYY x rryy)
R = round seeds, r = wrinkled seeds Y = yellow seeds, y = green seeds
What percent will have Round and Yellow seeds?
(RRYY x rryy)
R = round seeds, r = wrinkled seeds Y = yellow seeds, y = green seeds
What percent will have Round and Yellow seeds?
a)
0%
b)
25%
c)
50%
d)
100%
23.
What fraction of offspring from the cross AaBB x Aabb will be heterozygous for both traits?
a)
1/16
b)
4/16
c)
8/16
d)
3/16
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