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Home Q & A Board Biology-Related Homework Help General Biology
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ZOË95 ZOË95
wrote...
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11 years ago
Question 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 #1
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Question 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: Logic words? Two brown-eyed parents!

THUMBNAIL #2
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Question 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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Question 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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Question 24

Tall pea plants are dominant to dwarf pea plants. If no dwarf plants appeared after five generations of tall plants crossed with short plants, one can hypothesize that the tall plants:

A) were dominant
B) were homozygous
C) had the gene for dwarfness on the same chromosome
D) were heterozygous
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Question 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: 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 #3
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Question 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: 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 #4
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Question 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 #5
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Question 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 #5
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Question 29

A pea plant that is heterozgous tall and having homozygous green pods has the genotype and phenotype, respectively, of:
   
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:    
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 #6
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Question 30

The genotype of a parent generation in which a yellow heterozygous tall pea plant is pollinated with a dwarf heterozygous green plant is:
 
A) P1 =  Ttgg x ttGg
B) P1 =  Ttgg x ttgg
C) P1 =  Ttgg x ttGG
D) P1 =  TtGg x TtGg

HINT:
(a) TtGg  x  ttgg
(b) TtGg  x  TtGg (actually, any combination)
(c) ttGG  x  TTgg

If you start writting the tall trait first, continue this pattern all throughout the question.

THUMBNAIL #7
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wrote...
#1
11 years ago
Zoe, could you post the solutions to these questions. I really need them for my assignment. Frowning Face
wrote...
#2
11 years ago
16) 2, 1, 1, 2
homozygous means that both alleles are the same and when you use a punnett square, that only occurs when you mix n and n together.
the parents have genotypes of Nn and n so the offspring with one of those genotypes is genotype 1 (Nn).
heterozygous means that the alleles are different, so Nn is the only heterozygous genotype.
recessive phenotype means that the organism portrays the recessive allele (n) and that's only the case in nn because Nn would show the dominant trait.

17) D
use a punnett square. you known that 1 of the offspring has gray eyes so the 4th square you put gg and you know that both parents have brown eyes so both of their genotypes start with G, giving the 1st square a GG genotype.

18) B
BB and Bb would have brown phenotypes because they carry the dominant allele and bb would be white because both of their alleles are recessive.

19) D
each offspring gets a copy of 1 of the alleles so each sperm has a 50% chance of getting a W and a 50% of getting a w. so out of 4, 2 would get W and the other 2 would get w.

24) B
if the tall plant is heterozygous, the 4th square in a punnett square would be heterozygous recessive, making the plant a dwarf plant. but because we know that each of the offspring is tall, we know that the tall plant must be homozygous dominant.

25) B
a genotype is expressed with letters (RW, RR, WW) and the phenotype is the characteristic the organism has (pink, red, white). so automatically you know that B is the only possible answer.

26) A
use a punnet square to get the offspring of a black and white horse (all are BW) and then use another punnett square to get the offspring of BW and WW. 2 of the offspring have the genotype BW and the other 2 have WW. none of the offspring have the genotype BB, so none of them are black.

27) 5, 9, 4
the wording of this question seems to be wrong. it says that the parent generations are "heterozygous" but then in the question is says they are "pure", those terms contradict one another so I'm disregarding the word "pure". we know that the combination of red and white make pink so each allele must be dominant (RW), we also know that the parents are heterozygous so they must be RW. use a punnett square to fill in the offspring genotypes.

28) 1:2:1
use a punnett square to cross RW and RW

29) D
it tells you in the question what the phenotype is: heterozygous TALL and homozygous GREEN PODS. only 1 of the answers match up with the description in the question itself. also, the allele combination for heterozygous so you known it's Tt and the pod color is homozygous so it's GG.

30) A
because the 1st is heterozygous for tall, we known that the tall allele is dominant. and because the 2nd is heterozygous for green, we known that green must be dominant. the 1st heterozygous tall (Tt) and yellow (gg) and the 2nd plant is dwarf (tt) and heterozygous green (Gg).

Hopefully this helps someone Slight Smile
wrote...
#3
11 years ago
So 29 would be (a) then not (d)
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