4) In flowers, blue colour is dominant over white petal colour.
The genotype of a heterozygous blue flower and the phenotype of a white flower,
respectively, is:
A) white and black
B) Bb and bb
C) Bb and white
D) BB and bb
Hint for Question 4:
(a) bb, the same lower case
(b) BB, pure is the same as homozygous
(c) Bb, heterozygous means one of each
(d) bb, the recessive can only be written in one way
(e) BB, genotype is the letters, the gene formula
(f) blue; heterozygous is added to distract you
Locate the Logic word: phenotype
THUMBNAIL #1
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5) In a wild plant, white flowers are dominant while red is recessive.
The phenotype of a homozygous white flower and the genotype of a red flower,
respectively, is:
A) Ww and WW
B) WW and white
C) white and Ww
D) white and ww
Hint for Question 5:
(a) ww, any allele letters as lower case
(b) Ww, heterozygous; letters must be different
(c) WW, homozygous; letters must be the same
(d) red; phenotype!
(e) an impossibility: if red it has to be ww, homozygous; if it is heterozygous Ww, then it can't be red
THUMBNAIL #2
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6) If green is dominant to clear seed colour in a plant, then a cross between two plants that are heterozygous for green seed colour can be represented by:
A) P1 = Gg x GG
B) P1 = GG x GG
C) P1 = Gg x Gg
D) P1 = GG x gg
Hint for Question 6:
(a) P1 = ss x Ss
(b) P1 = BB x bb
(c) P1 = bb x Bb
Information:
Once symbols are assigned, the next step is to use the information to correctly form the gene formula (genotype) of the parents. Please practice making the following parents. Before you attempt the multiple choice question check the Hint to evaluate your answers.
a) Make the parent generation of the homozygous tall tree pollinated by a heterozygous short tree. Shortness is dominant.
b) A yellow flower is recessive. Write the parent generation in a cross between homozygous blue and a yellow flower.
c) Brown hair is dominant to red hair. Make the parent generation between a red haired mother and a heterozygous brown haired father.
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9) When conducting his research into the genetics of garden peas, Gregor Mendel crossed true-breeding, yellow-seeded plants with true-breeding, green-seeded plants. He observed that all the F1 generation had yellow seeds.
In the F1 generation, the allele for yellow seeds would have been found in:
A) about 50% of the female gametes and 50% of the male gametes produced by F1 plants
B) all the female gametes and none of the male gametes produced by the F1 plants
C) none of the male or female gametes produced by the F1 plants
D) all the male gametes and none of the female gametes produced by the F1 plants
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11) If (Q) represents the gene for a dominant trait and (q) is the recessive allele, and if Qq mates with qq, then:
A) 50% of the offspring will be recessive
B) all offspring will be of the dominant trait
C) all the offspring will be of the recessive trait
D) 75% of the offspring will be of the dominant trait
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12) If two offspring result from the cross Zz x Zz, what is the probability that both will have the same genotype ZZ?
A) 1/4
B) 1/8
C) 1/16
D) 1/2
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13) If a short pea plant (recessive) is self-pollinating, then the phenotype of the offspring will be:
A) 0% short
B) 3:1; 3 tall:1 short
C) 50 / 50; half short / half tall
D) 100% short
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In birds, black feathers are dominant over white feathers. A black hen is mated to a black rooster. Her eggs were saved. Forty-two chicks hatched in an incubator: thirty-one were black and eleven were white. The gene formulas for the two parents and their offspring are:
A) P1 = Bb x bb ; F1 genotypes: Bb, bb, Bb, bb
B) P1 = BB x Bb_; F1 genotypes: BB, BB, BB
C) P1 = BB x bb ; F1 genotypes: BB, bb, Bb
D) P1 = Bb x Bb ; F1 genotypes: BB, bb, Bb
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15) Numerical Response
In the crossing genotype 1 with genotype 2, in what proportion of offspring would the recessive phenotype appear?
Record your answer as a value from 0 to 1, rounded to two significant digits.
Answer:
THUMBNAIL #3
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16) Numerical Response
Choose the correct genotype number for each description of an offspring.
Genotype number: ____________ ____________ ____________ ____________
Offspring Description: Homozygous Genotype similar to Heterozygous Recessive
genotype one of the parents genotype phenotype
THUMBNAIL #4
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17) Gray eyes are inherited as a recessive trait. Two brown-eyed individuals had a gray-eyed offspring. Which of the above crosses would result from the genotype of the parents?
A) A.
B) B.
C) C.
D) D.
Hint for Question 17:
Logic words?
Two
brown-eyed parents!
THUMBNAIL #5
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18) In mice with complete dominance in coat colour, brown (B) is dominant to white (b). The genotypes BB, Bb, and bb represent:
A) two genotypes
B) two phenotypes
C) three phenotypes
D) one phenotype
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19) Mendel's Law of Segregation separates alleles in forming gametes. Beginning with the genotype (Ww), how would these genes be distributed among four sperm cells?
A) all W or all w
B) many ways as a matter of chance
C) WW, Ww, Ww, ww
D) two containing w and two containing W
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25) If incomplete dominance exists between red and white snapdragons, then the correct genotype of a pink flower and the phenotype of a red flowered-parent,
respectively, is:
A) RW:RR
B) RW:parent → red
C) pink:RR
D) WW:RR
Hint for Question 25:
After reading (sr) gen 2, did you change the format in how you assigned symbols?
(a) BB (b) green (c) YY (d) YB or BY
(e) No,the individual would be green (BY)
If you wrote the Y first in YB you would continue to do so all through the Punnett square.
BY would be equally correct, but then the B would be written first for all green BY's in the question.
THUMBNAIL #6
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26) If there is a lack of dominance among black and white horses, then the percentage that will be black if a white horse is mated with a gray horse is:
A) 0%
B) 75%
C) 50%
D) 25%
Hint for Question 26:
Note that in codominance or incomplete dominance, it is not necessary to use homozygous for black or white. Since black can only be BB, a heterozygous BW would have the phenotype gray.
THUMBNAIL #7
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27) In a case of incomplete dominance, pink flowers are produced from the mating of pure four-o'clock plants.
Numerical Response
Choose the correct allele combination number from the legend for each of the folowing.
Number : ___________ ___________ ___________
genotype 4 allele 2 genotype 3
THUMBNAIL #8
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28) In a case of incomplete dominance, pink flowers are produced from the mating of pure four-o'clock plants.
Numerical Response
If a first filial cross mates genotype 3 x genotype 2, then the phenotypic ratio is
Answer: ______ : ______ : ______
white pink red
THUMBNAIL #9
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29) A pea plant that is heterozgous tall and having homozygous green pods has the genotype and phenotype, respectively, of
Question 29 options:
A) TTGG: height → dwarf; pod colour → green
B) TTGg: height → tall; pod colour → yellow
C) ttgg: height dwarf; pod colour → yellow
(This is a true description of ttgg, but does it answer the question?)
D) TtGG: height → tall; pod colour → green
Hint for Question 29:
If the questions represent an individual parent, each genotype should have
two alleles for the first trait, and two alleles for the second trait [4 letters in total].
(a)
TTGG; if the symbol
T for tallness is written first, then continue with
T or t's first all throughout the question.
(b)
TtGg; both traits exhibit the heterozygous condition
(c)
ttgg; even if described as yellow dwarf, it is written ttgg as the tall trait has been indicated to be written first.
(d)
TTGG;
even though green is described first, TTGG
(e)
Ttyy(f)
ttGg; are you getting better at forming the genotype in dihybrid inheritance?
(g)
ttgg (homozygous recessive); used to test if the other parent is homozygous or heterozygous for a trait.
THUMBNAIL #10
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