Question 1

Two parents with widow's peak and short fingers have a child with a continuous hairline and long fingers.What are the genotypes of the two parents? (W = widow's peak; S = short fingers)
A) WWSS x wwss
B) wwss x WWSS
C) WWss x WWss
D) WwSs x WwSs
E) WwSs x wwss

Accepted Answer

If the child's phenotype is different for both dominant traits from that of the parents then the parents must be heterozygous for both traits. This would mean that they possess the recessive alleles to pass on to their offspring.Bloom's

Question 2

Which of the following crosses will yield four phenotypes in a ratio of 9:3:3:1?
A) RRYY x RRYY
B) RRYY x rryy
C) RrYy x RrYy
D) RrYy x rryy
E) rryy x rryy

Accepted Answer

For a dihybrid cross the parents would need to be heterozygous for both of two traits. RrYy x RrYy would produce RRYY, RRYy, RRyy, RrYY, Rryy, rrYY, rrYy, or rryy as genotypes the phenotypes of these would be 9 dominant for both traits to 3 recessive for the first trait and dominant for the second trait to 3 dominant for the first trait and recessive for the second and 1 recessive for both traits.Bloom's

Question 3

What are the chances that two individuals with wavy hair (an incomplete trait)will have a curly-haired child? Curly hair and straight hair exhibit incomplete dominance.
A) none
B) 25%
C) 50%
D) 75%
E) 100%

Accepted Answer

Given that WW = curly hair, ww = straight hair and Ww = wavy hair; if both parents have wavy hair then their genotype is Ww. A cross of Ww x Ww would yield a 25% chance for WW (curly) a 50% chance for Ww (wavy) and a 25% chance for ww (straight) the ratio would be 1:2:1.Bloom's

Question 4

In horses,black coat (B)is dominant over brown coat (b),and being a trotter (T)is dominant over being a pacer (t).A black horse who is a pacer is crossed with a brown horse who is a trotter.The offspring is a brown pacer.What is the genotype of the brown trotter parent?
A) BBTT
B) BbTt
C) Bbtt
D) bbTt
E) bbtt

Accepted Answer

The cross give is B__tt x bbT___ with the blanks showing that the genotype for the second allele is not known because the dominant phenotype is seen. To produce a brown pacer (bbtt) the parents would need to have the recessive alelle for both traits so their genotypes would be Bbtt and bbTt.Bloom's

Question 5

If one parent has type B blood and the other has type AB,what type might the child have?
A)A or O only
B)B or O only
C)A or B only
D)A,B,AB only
E)A,B,O,or AB

Accepted Answer

The answer of If one parent has type B blood...

Question 6

Mr.X has been accused by Miss Z of fathering her child.The child's blood type is AB and Miss Z's blood type is A.Mr.X could NOT be the father if his blood type is
A)only A.
B)only B.
C)AB.
D)only O.
E)either A or O.

Accepted Answer

Bloom's

Question 7

The inheritance of blood types in humans can be explained by
A) simple Mendelian genetics (one gene pair).
B) linked genes.
C) simple dominance.
D) multiple alleles.
E) polygenic inheritance.

Accepted Answer

Blood types in humans involve multiple alleles that are codominant. Simple Mendelian genetics would indicate a simple dominant/recessive relationship and polygenetic inheritance would mean that there were multiple genes on different chromosomes. Linked genes are passed as a group where as blood types are a single gene with multiple alleles.Bloom's

Question 8

In pea plants,the gene for round seed (R)is dominant,and wrinkled seeds (r)are recessive.The endosperm of the pea is also either starchy,a dominant gene (S),or waxy (s).What CANNOT be said of a cross between two parents that are heterozygous for both traits?
A) It is possible to secure offspring that are homozygous for both dominant genes.
B) It is possible to secure offspring that are homozygous for both recessive genes.
C) It is possible to secure offspring that are homozygous for one dominant gene such as round seed and homozygous recessive for the other recessive waxy gene.
D) It is possible to secure offspring that are heterozygous for both genes.
E) It is impossible to secure offspring that are homozygous for both dominant genes.

Accepted Answer

A dihybrid cross would be RrSs x RrSs and this would produce RRSS, RRSs, RRss, RrSS, Rrss, rrSS, rrSs, or rrss as genotypes the phenotypes of these would be 9 dominant for both traits ( roun and starchy seeds) to 3 recessive for the first trait and dominant for the second trait (wrinkled and starchy seeds) to 3 dominant for the first trait and recessive for the second (round and waxy seeds) and 1 recessive for both traits (wrinkled and waxy seeds).Bloom's

Question 9

In which kind of cross could you expect to find of ratios of 1:1:1:1 among the offspring?
A) monohybrid cross
B) dihybrid cross
C) one trait test cross
D) two trait test cross
E) no cross will yield offspring in 1:1:1:1 ratios

Accepted Answer

The 1:1:1:1 ratio is expected when an organism heterozygous for two traits is crossed with an organism that is homozygous recessive for both traits. This is also a test cross that is frequently used to determine the genotype of organisms that show the dominant traits.Bloom's

Question 10

In guinea pigs,B = black,b = brown,S = short hair,s = long hair.A heterozygous black,shorthaired animal reproduces with a brown,longhaired animal.What is the expected phenotypic ratio of the offspring?
A) 1 black short hair, 1 black long hair, 1 brown short hair, 1 brown long hair
B) 3 black short hair, 1 brown long hair, 3 black long hair, 1 brown short hair
C) 9 black short hair, 3 black long hair, 3 brown long hair, 1 brown short hair
D) 9 black short hair, 3 black long hair, 3 brown short hair, 1 brown long hair
E) 9 black short hair, 6 black long hair, 3 brown long hair, 1 brown short hair.

Accepted Answer

The cross of BbSs x bbss would yield BbSs, bbSs, Bbss and bbss as possible genotypes. The ratio of phenotypes would be 1 black short hair to 1 black long hair to 1 brown short hair to 1 brown long hair.Bloom's

Question 11

Which of the following crosses will yield 4 phenotypes in the 1:1:1:1 ratio?
A) RRYY x RRYY
B) RRYY x rryy
C) RrYy x RrYy
D) RrYy x rryy
E) rryy x rryy

Accepted Answer

A cross between an individual that is heterozygous for two traits and one that is homozygous recessive for those two traits would yield a ratio of 1 dominant for both traits to 1 dominant for the first trait and recessive for the second to 1 recessive for the first trait and dominant for the second to 1 recessive for both traits.Bloom's

Question 12

A blue-eyed,left-handed woman marries a brown-eyed,right-handed man who is heterozygous for both of his traits.If blue eyes and left-handedness are recessive,how many different phenotypes are possible in their children?
A) one
B) two
C) three
D) four
E) five

Accepted Answer

If B = brown eyes and b = blue eyes and R = right handed while r = left handed then the mother's genotype is bbrr and the father is BbRr. A cross between an individual that is homozygous recessive for two traits and one that is heterozygous for those two traits then there would be a ratio of 1 dominant for both traits to 1 dominant for the first trait and recessive for the second to 1 recessive for the first trait and dominant for the second to 1 recessive for both traits, yielding four different phenotypes.Bloom's

Question 13

The four o'clock flower is an example of incomplete dominance: R = red,r = white,and Rr = pink.If two hybrids are crossed,what are the chances that an offspring will have pink flowers?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%

Accepted Answer

Rr x Rr will produce RR, Rr and rr flowers since the parents have both alleles they have the abliity to pass both on to their offspring. This is the cross that Mendel made when he crossed the F₂ with itself. We can calculate that the offspring's genotypes will be 25% RR, 50% Rr and 25% rr.Bloom's

Question 14

A cross is made between two parents with genotypes AaBB and aabb.If there are 32 offspring,how many of them would be expected to exhibit both dominant characteristics?
A) 32
B) 24
C) 16
D) 8
E) 0

Accepted Answer

A cross between AaBB x aabb would produce AaBb and aaBb as possible genotypes. This means that there is a 50% chance that the offspring would be dominant for both traits. If 32 offspring are born then half of those, 16, would be expected to have this phenotype.Bloom's

Question 15

Which of the following is an example of the blending of phenotypes?
A) codominance
B) sex-linked inheritance
C) polygenic inheritance
D) simple Mendelian inheritance
E) incomplete dominance

Accepted Answer

Blending is when the offspring of homozygous parents do not show the phenotype of either of the parents but rather a unique third phenotype.Bloom's

Question 16

A man with blood type AB could NOT be the father of a child with
A)blood type A.
B)blood type B.
C)blood type AB.
D)blood type O.
E)A man with blood type AB can have a child of any blood type.

Accepted Answer

If the father has blood type AB he could father a child with blood types AB, A or B; the only blood type that could be eliminated would be O.Bloom's

Question 17

In humans,aniridia,a type of blindness,is due to a dominant allele A.Migraine headaches are due to another dominant allele M.If a man who suffers from both conditions (AaMm)marries a woman who suffers from both (AaMm),what are the chances of an offspring having both traits?
A)9/16
B)3/16
C)1/2
D)1/16
E)4/16

Accepted Answer

The man is AaMm and he marries a woman who is AaMm. This is a dihybrid cross and would produce AAMM, AAMm, AAmm, AaMM, Aamm, aaMM, aaMm, or aamm as genotypes the phenotypes of these would be 9 dominant for both traits to 3 recessive for the first trait and dominant for the second trait to 3 dominant for the first trait and recessive for the second and 1 recessive for both traits.Bloom's

Question 18

A man with widow's peak (dominant)who can curl his tongue (dominant)has a child who has a continuous hairline and cannot curl the tongue.What is the genotype of the father?
A) WWTT
B) wwtt
C) WwTt
D) Wwtt
E) WT

Accepted Answer

Given that W = widow's peak and w = continuous hair line and T = roll tongue while t = cannot roll tongue then if the child's phenotype is different for both dominant traits from that of the parent then the parent must be heterozygous for both traits. This would mean that he possess the recessive alleles to pass on to his offspring. His genotype would be WwTt.Bloom's

Question 19

In guinea pigs,B = black,b = brown,S = short hair,s = long hair.Two heterozygous individuals reproduce.The expected results are
A) 9 black long hair, 3 black short hair, 3 brown long hair, 1 brown short hair.
B) 9 black short hair, 6 black long hair, 3 brown long hair, 1 black short hair.
C) 9 black short hair, 3 black long hair, 3 brown short hair, 1 brown long hair.
D) 9 brown short hair, 3 black long hair, 3 brown long hair, 1 black short hair.
E) 1 black short hair, 1 black long hair, 1 brown short hair, 1 brown long hair.

Accepted Answer

For a dihybrid cross the parents would need to be heterozygous for both of two traits. BbSs x BbSs would produce BBSS, BBSs, BBss, BbSS, Bbss, bbSS, bbSs, or bbss as genotypes the phenotypes of these would be 9 dominant for both traits (black and short hair) to 3 recessive for the first trait and dominant for the second trait (brown with short hair) to 3 dominant for the first trait and recessive for the second (black with long hair) and 1 recessive for both traits (brown with long hair).Bloom's

Question 20

In which kind of cross would you expect to find a ratio of 9:3:3:1 among the F₂ offspring? A) monohybrid cross B) dihybrid cross C) test cross D) homozygous recessive x homozygous dominant E) homozygous dominant x heterozygous

Accepted Answer

For a dihybrid cross the parents would need to be heterozygous for both of two traits. Example: RrYy x RrYy would produce RRYY, RRYy, RRyy, RrYY, Rryy, rrYY, rrYy, or rryy as genotypes the phenotypes of these would be 9 dominant for both traits to 3 recessive for the first trait and dominant for the second trait to 3 dominant for the first trait and recessive for the second and 1 recessive for both traits.Bloom's

Quiz 23: Patterns of Gene Inheritance

Two parents with widow's peak and short fingers have a child with a continuous hairline and long fingers.What are the genotypes of the two parents? (W = widow's peak; S = short fingers)

Which of the following crosses will yield four phenotypes in a ratio of 9:3:3:1?

What are the chances that two individuals with wavy hair (an incomplete trait) will have a curly-haired child? Curly hair and straight hair exhibit incomplete dominance.

In horses,black coat (B) is dominant over brown coat (b) ,and being a trotter (T) is dominant over being a pacer (t) .A black horse who is a pacer is crossed with a brown horse who is a trotter.The offspring is a brown pacer.What is the genotype of the brown trotter parent?

If one parent has type B blood and the other has type AB,what type might the child have?

Mr.X has been accused by Miss Z of fathering her child.The child's blood type is AB and Miss Z's blood type is A.Mr.X could NOT be the father if his blood type is

The inheritance of blood types in humans can be explained by

In pea plants,the gene for round seed (R) is dominant,and wrinkled seeds (r) are recessive.The endosperm of the pea is also either starchy,a dominant gene (S) ,or waxy (s) .What CANNOT be said of a cross between two parents that are heterozygous for both traits?

In which kind of cross could you expect to find of ratios of 1:1:1:1 among the offspring?

In guinea pigs,B = black,b = brown,S = short hair,s = long hair.A heterozygous black,shorthaired animal reproduces with a brown,longhaired animal.What is the expected phenotypic ratio of the offspring?

Which of the following crosses will yield 4 phenotypes in the 1:1:1:1 ratio?

A blue-eyed,left-handed woman marries a brown-eyed,right-handed man who is heterozygous for both of his traits.If blue eyes and left-handedness are recessive,how many different phenotypes are possible in their children?

The four o'clock flower is an example of incomplete dominance: R = red,r = white,and Rr = pink.If two hybrids are crossed,what are the chances that an offspring will have pink flowers?

A cross is made between two parents with genotypes AaBB and aabb.If there are 32 offspring,how many of them would be expected to exhibit both dominant characteristics?

Which of the following is an example of the blending of phenotypes?

A man with blood type AB could NOT be the father of a child with

In humans,aniridia,a type of blindness,is due to a dominant allele A.Migraine headaches are due to another dominant allele M.If a man who suffers from both conditions (AaMm) marries a woman who suffers from both (AaMm) ,what are the chances of an offspring having both traits?

A man with widow's peak (dominant) who can curl his tongue (dominant) has a child who has a continuous hairline and cannot curl the tongue.What is the genotype of the father?

In guinea pigs,B = black,b = brown,S = short hair,s = long hair.Two heterozygous individuals reproduce.The expected results are

In which kind of cross would you expect to find a ratio of 9:3:3:1 among the F₂ offspring?