Question 1

Which of the following is NOT correct concerning the law of independent assortment?
A) It is based upon the process of meiosis.
B) Each pair of factors separates independently.
C) All possible combinations of factors can occur in the gametes.
D) It was the second law of heredity established by Mendel.
E) It follows the observation that all maternal chromosomes end up in the egg.

Accepted Answer

Independent assortment occurs because homologous chromosomes are separated during meiosis I. This separated the alleles that are contained on these chromosomes. Because the probability of either allele ending up in the gamete is equal then all possible combinations of alleles from the parents is equally likely. Mendel discovered this law through his experiments using monohybrid crosses. Only half of the chromosomes from the mother end up in the egg.Bloom's

Question 2

If pairs of factors separate independently of other pairs of factors,you are dealing with the
A) law of segregation.
B) law of independent assortment.
C) law of dominance.
D) multiple alleles.
E) codominance.

Accepted Answer

The law of independent assortment addresses the fact that unlinked genes will be inherited independent of how the other is inherited. The law of segregation addresses the fact that two alleles for the same gene are separated and only one is inherited from each parent. The law of dominance says that given two alleles for the same trait one can be dominant over the other; if both are present in the organism then only one will be shown and it will always be the same one. With codominance both of the alleles are expressed equally.Bloom's

Question 3

Some plants fail to produce chlorophyll,and this trait appears to be recessive.Many plants also self-pollinate.If we locate a pea plant that is heterozygous for this trait,self-pollinate it and harvest seeds,what are the likely phenotypes of these seeds when they germinate?
A) All will be green with chlorophyll since that is the dominant trait.
B) All will be white and lack chlorophyll since this is self-pollinated.
C) About 1/2 will be green and one-half white since that is the distribution of the genes in the parents.
D) About 1/4 will be white and three-fourths green since it is similar to a monohybrid cross.
E) About 1/4 will be green and three-fourths white since it is similar to a monohybrid cross.

Accepted Answer

Where g = no chlorophyll and G = chlorophyll, a heterozygote would have the genotype Gg. If the plant self pollinates then the genotype for ova production is Gg and the genotype for pollen production would be Gg. This monohybrid cross would yield the 1:2:1 ratio (25%, 50%, 25%) typical of such a cross.Bloom's

Question 4

Computer simulations are sometimes used to demonstrate the outcome of monohybrid fruit fly crosses,where a student can run generation after generation of fruit flies with 100 offspring produced each generation,half male and half female,and a 3-to-1 phenotype ratio (or 75 to 25)in the F₁ generation.Compared to real genetics results,which of the following is NOT true? A) Rarely would exactly 100 fly offspring be produced or survive. B) An exact balance between males and females would be rare. C) A precise 3-to-1 ratio would be uncommon. D) It is usually possible to have exactly 100 offspring, half male and half female, and a 3-to-1 phenotypic outcome in real life. E) Ratios and outcomes will not be exact but will be very close to the expected numbers.

Accepted Answer

In real life it is very uncommon for exact numbers or ratios to be generated in an experiment. With a computer simulation it is possible to obtain exactly the expected ratios or numbers since the computer model is designed by humans who program in these expected values.Bloom's

Question 5

If an individual with a dominant phenotype is crossed with an individual with a recessive phenotype,4 of their 9 offspring show the recessive phenotype.What is the genotype of the dominant parent?
A) AA
B) Aa
C) aa
D) either AA or Aa
E) either Aa or aa

Accepted Answer

If one of the parents has a recessive phenotype then the other parent must have a dominant allele in order to have a child with the dominant phenotype.Bloom's

Question 6

In cattle,consider that the lack of horns is dominant to the presence of horns.A bull without horns is crossed with a horned cow.Of four offspring,one is horned and the other three lack horns.The bull is
A) homozygous dominant.
B) homozygous recessive.
C) heterozygous.
D) heterozygous dominant.
E) heterozygous recessive.

Accepted Answer

Given that H=lack of horns and h=horns, if the mother is horned and the father is hornless and calves with both phenotypes are produced by the cross, then female who has has horns must be (hh) and the bull must be Hh. Since the mother only has one allelic choice to give (h) then the father must have both alleles.Bloom's

Question 7

Unattached earlobes (EE or Ee)are dominant over attached earlobes (ee).A couple both have unattached earlobes.Both notice that one of their parents on both sides have attached earlobes (ee).Therefore,they correctly assume that they are carriers for attached earlobes (Ee).If the couple proceeds to have four children,then
A) they can be certain that three will be heterozygous and one homozygous recessive.
B) if the first three are heterozygous, the fourth must be homozygous recessive.
C) the children must repeat the grandparents' genotype (Ee).
D) all children must have unattached earlobes since both parents possess the dominant gene for it.
E) two heterozygous, one homozygous dominant, and one homozygous recessive is a likely outcome, but all heterozygous, or two, three or all four homozygous is also possible.

Accepted Answer

The chance of any one child possessing a particular phenotype does not depend on the genotype of any sibling. The parents still have the same genotype and the eggs and sperm are created from germ cells which contain both allele choices that the parents have. Therefore, the probablitiy is the same for each child to have any one particular genotype. In this case the probabilities would be 75% chance of unattached earlobes and a 25% chance of attached earlobes.Bloom's

Question 8

In 1940,two researchers named Weiner and Landsteiner discovered that about 85 percent of the human population sampled possessed a blood cell protein called the Rh factor.The presence of Rh factor was labeled Rh positive (Rh⁺)and was found to be dominant over the absence of the blood factor (Rh^-).Under normal Mendelian inheritance,which of the following statements is/are false? A) Two Rh⁺ parents could have an Rh^- child. B) Two Rh^- parents could have an Rh⁺ child. C) An Rh^- child would require that both parents be carriers of at least one Rh^- allele. D) It is possible with just one pair of parents to have children where some siblings are Rh^- and some are Rh⁺. E) Two Rh⁺ parents can have Rh^- or Rh ⁺ children.

Accepted Answer

Rh⁺Rh^- would still show an Rh⁺ phenotype so, if both parents were heterozygous, they could have an Rh^-child. Both parents would need to carry the Rh^- allele. Since the parents also have the Rh⁺ allele they could also have children of this phenotype. If both parents have the Rh-phenotype then their genotype must be Rh^- Rh^- and it would not be possible for them to have a child with the Rh⁺ phenotype.Bloom's

Question 9

Since each child of two heterozygous parents has a 1/4 chance of receiving a recessive trait from each parent,
A) if the first child is phenotypically recessive, then the next child must be phenotypically dominant.
B) if the first child is phenotypically recessive, then the next child has a 3/4 chance of being phenotypically recessive.
C) if the first child is phenotypically recessive, then the next child has a 1/2 chance of being phenotypically recessive.
D) if the first child is phenotypically recessive, then the next child has a 1/4 chance of being phenotypically dominant.
E) no matter what the first child's phenotype, the next child will have a 1/4 chance of being phenotypically recessive.

Accepted Answer

The chance of any one child possessing a particular phenotype does not depend on the genotype of any sibling. The parents still have the same genotype and the eggs and sperm are created from germ cells which contain both allele choices that the parents have. So the probablitiy is the same for each child to have any one particular genotype.Bloom's

Question 10

Both parents are heterozygous for Tay-Sachs disease (an autosomal recessive disorder).Three children in a row were born with Tay-Sachs disease.What is the chance that a fourth child will have Tay-Sachs disease?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%

Accepted Answer

The chance of any one child possessing a particular phenotype does not depend on the genotype of any sibling. The parents still have the same genotype and the eggs and sperm are created from germ cells which contain both allele choices that the parents have. Therefore, the probabiltiy is the same for each child to have any one particular genotype. With two heterozygotes this would mean a 75% chance of a normal phenotype and a 25% chance of a recessive phenotype.Bloom's

Question 11

If the probability of event A is 3/4 and the probability of event B is 1/4,then the probability of both A and B occurring at the same time is
A) 3/4.
B) 1/4.
C) 1 or absolute certainty.
D) 1/2.
E) 3/16.

Accepted Answer

The multiplication rule applies to probability, that is we multiply the probability of one event by the probability of the other event. In this case A x B = 3/4 x 1/4 = 3/16.Bloom's

Question 12

If the genotype of an organism is YySsTt,then yST would represent
A) the genotype of the offspring.
B) a possible phenotype of the offspring.
C) a gamete of the parent.
D) a possible zygote.
E) the P generation.

Accepted Answer

Given three genes Y, S and T, then YySsTt could represent the genotype of an individual who is heterozygous for all three traits and yST could represent a possible combination of the alleles that could be produced in a gamete. If the genotype of either the offspring or a parent were to be listed then there would need to be two alleles for each gene since an zygote or an organism would have the full complement of chromosomes.Bloom's

Question 13

A cross that involves two traits is called a ______ cross.
A) monohybrid
B) dihybrid
C) Punnett
D) test
E) homozygous

Accepted Answer

A monohybrid cross follows one trait while a dihybrid cross follows two traits that assort independently of one another. A Punnett square may be used to determine the out come of either of this types of crosses.Bloom's

Question 14

In humans,assume red hair is recessive to dark hair.What are the chances of a dark-haired couple having a red-haired child if each had a red-haired parent?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%

Accepted Answer

Given that D = dark hair and d = red hair, if both of the parents, who have dark hair, each had a parent with red hair, which is recessive, then both must be heterozygous for hair color (Dd). This would be a monohybrid cross and the chance of a homozygous recessive from this match would be 25%.Bloom's

Question 15

In which kind of cross would you expect to have your actual results match the predicted results 100% of the time the cross is made?
A) monohybrid cross
B) dihybrid cross
C) test cross
D) independent assortment cross
E) crosses of differing genotypes result in differing answers-no cross will produce expected ratios every time

Accepted Answer

Living systems seldom behave exactly as expected. While the ratios will be very close to what is predicted that will not necessarily be exactly what is expected.Bloom's

Question 16

Which of the following would be considered a heterozygous dihybrid genotype?
A) Ww
B) sS
C) WWSs
D) WWss
E) WwSs

Accepted Answer

A dihybrid would mean that there were two different genes involved. WwSs would be heterozygous for both traits. WWss or WWSs would not be heterozygous and sS or Ww would not show the information for two traits.Bloom's

Question 17

A pheasant breeder starts with two birds in the P generation,one of which is AA and the other is aa.If he takes two of the birds from the F₁ generation and breeds them together,what can he expect in his F₂ offspring? A) AA and Aa. B) Aa and aa. C) AA, Aa, and aa. D) AA only. E) Aa only.

Accepted Answer

If the parents (P generation) are AA and aa, then all of the F₁offspring must be Aa since they will get one allele from each parent, and each parent only has one allele choice to give. When the F1 Aa individuals are crossed, the F₂ generation will have the genotypes AA, Aa, and aa.Bloom's

Question 18

If an albino (recessive)woman is married to a man with normal coloring and they have an albino child,what was the genotype of the man?
A) homozygous dominant
B) heterozygous
C) homozygous recessive
D) incomplete dominance
E) heterozygous dominant

Accepted Answer

Given that A = normal coloring and a = albino coloring an albino recessive mother is aa and with normal coloring the father can either be AA or Aa. If the father is AA then all of the children would have normal coloring. If the father is Aa then there is a 50% probability of a child with normal coloring and a 50% probability of an albino child. Therefore the father must be heterozygous for the trait.Bloom's

Question 19

A woman who can roll her tongue (presumably dominant)is married to a man who cannot.Two of their four children can roll their tongues and two cannot.If A = roll tongue and a = cannot roll tongue,then what is the genotype of the parents?
A) woman Aa; man Aa
B) woman AA; man aa
C) woman Aa; man AA
D) woman Aa; man aa
E) woman aa; man aa

Accepted Answer

Since 'cannot roll tongue' is recessive the father must be aa. If some of the children can roll their tongue and some others cannot then the mother must have had both alleles (heterozygous) Aa.Bloom's

Question 20

Which of the following crosses will yield only homozygous recessive offspring?
A) RRYY x RRYY
B) RRYY x rryy
C) RrYy x rryy
D) rryy x rryy
E) RrYy x RrYy

Accepted Answer

To produce only homozygous recessive offspring only homozygous recessive parents could be mated. If any dominant alleles were present in the parents there would be a probability that they would be present in the offspring.Bloom's

Quiz 23: Patterns of Gene Inheritance

Which of the following is NOT correct concerning the law of independent assortment?

If pairs of factors separate independently of other pairs of factors,you are dealing with the

Some plants fail to produce chlorophyll,and this trait appears to be recessive.Many plants also self-pollinate.If we locate a pea plant that is heterozygous for this trait,self-pollinate it and harvest seeds,what are the likely phenotypes of these seeds when they germinate?

If an individual with a dominant phenotype is crossed with an individual with a recessive phenotype,4 of their 9 offspring show the recessive phenotype.What is the genotype of the dominant parent?

In cattle,consider that the lack of horns is dominant to the presence of horns.A bull without horns is crossed with a horned cow.Of four offspring,one is horned and the other three lack horns.The bull is

Since each child of two heterozygous parents has a 1/4 chance of receiving a recessive trait from each parent,

Both parents are heterozygous for Tay-Sachs disease (an autosomal recessive disorder) .Three children in a row were born with Tay-Sachs disease.What is the chance that a fourth child will have Tay-Sachs disease?

If the probability of event A is 3/4 and the probability of event B is 1/4,then the probability of both A and B occurring at the same time is

If the genotype of an organism is YySsTt,then yST would represent

A cross that involves two traits is called a ______ cross.

In humans,assume red hair is recessive to dark hair.What are the chances of a dark-haired couple having a red-haired child if each had a red-haired parent?

In which kind of cross would you expect to have your actual results match the predicted results 100% of the time the cross is made?

Which of the following would be considered a heterozygous dihybrid genotype?

A pheasant breeder starts with two birds in the P generation,one of which is AA and the other is aa.If he takes two of the birds from the F₁ generation and breeds them together,what can he expect in his F₂ offspring?

If an albino (recessive) woman is married to a man with normal coloring and they have an albino child,what was the genotype of the man?

A woman who can roll her tongue (presumably dominant) is married to a man who cannot.Two of their four children can roll their tongues and two cannot.If A = roll tongue and a = cannot roll tongue,then what is the genotype of the parents?

Which of the following crosses will yield only homozygous recessive offspring?