Download Biology 179 - MSU Billings

Science 103: Practice Questions for Exam 3
SHORT ANSWER QUESTIONS:
1. An individual lost his father to Huntington’s disease. His mother is healthy and in her 60’s.
(a) What is the probability that the individual will develop Huntington’s disease?
(b) What is the probability that his children will be carriers of the disease?
2. A form of hemophilia is determined by a gene on the X chromosome in humans. Assume that a phenotypically normal woman
whose father had hemophila is married to a normal man. What is the probability that their first daughter will have hemophilia?
Draw a pedigree (showing all 3 generations) and explain your answer.
1
3. The following pedigree is concerned with an inherited dental abnormality, amelogenesis imperfecta. Normal individuals and
carriers are represented by the white symbols.
(a) What mode of inheritance best accounts for the transmission of this trait?
(b) Write all the genotypes of the family members (that are possible to deduce) on the pedigree.
4. When Mendel used pollen from a dwarf strain of peas to fertilize the flowers of a true-breeding tall line, and planted the
resulting seeds, only tall plants grew from them. When he crossed these hybrids to the dwarf plants he observed among the
offspring 87 tall and 79 dwarf plants. Diagram the cross.
2
5. For each of the pedigrees below, deduce the inheritance of the condition. Then, for each pedigree label all the individuals (that
are possible to deduce) with their genotype.
6. In dogs, dark coat color is dominant over albino and short hair is dominant over long hair. A double heterozygote is crossed
with a dog that is heterozygous for coat color but homozygous recessive for hair length.
(a) What genotypic and phenotypic ratios will be obtained? (show all working).
3
(b) How could you determine the genotype of a dark short-haired dog? (show all working).
7. In watermelons, the genes for green color and short shape are dominant over their alleles for striped color and long shape.
(a) Suppose a plant with long-striped fruit is crossed with a plant heterozygous for both these characters. What phenotypes would
be produced and in what ratios?
4
(b) Repeat part (a) using a cross between 2 watermelons that are heterozygous for both characters.
8. If the litter resulting from the mating of 2 short-tailed cats contains 3 kittens without tails, 2 with long tails, and 6 with short
tails, what would be the simplest way of explaining the inheritance of tail length in these cats? (SHOW ALL THE GENOTYPES
AND PHENOTYPES OF PARENTS AND PROGENY AND EXPLAIN WHAT TYPE OF INHERITANCE THIS IS).
5
9. Male house cats are either black or orange; females are black, orange, or calico.
(a) If these coat-color phenotypes are governed by a sex-linked gene, how can these observations be explained?
(b) What phenotypic and genotypic ratios would be expected for the F1 of a cross between an orange female and a black male.
(c) Repeat part (b) for the reciprocal of the cross described there.
(d) Half of the females produced by a certain kind of mating are calico, and half are black. Half of the males are orange and half
are black. What colors are the parents in this kind of mating?
6
MULTIPLE CHOICE QUESTIONS:
1.
To ensure true-breeding parental plants Mendel:
(A) self fertilized each variety for several generations.
(B) crossed 2 varieties with alternate traits.
(C) self-fertilized the F1 generation.
(D) A and B.
(E) none of the above.
2.
When Mendel crossed 2 varieties of true-breeding plants the F1 generation were:
(A) 100% heterozygous.
(B) 100% homozygous.
(C) 50% heterozygous.
(D) 75% heterozygous.
(E) none of the above.
The following statement is for questions 3 and 4:
In a particular breed of cats, black (B) is dominant to white (b). A black heterozygous cat is crossed with a black homozygous cat.
3.
The genotypic ratio is:
(A) 1:1 homozygous to heterozygous.
(B) 3:1 heterozygous to homozygous.
(C) All heterozygous.
(D) All black.
(E) 1:1 black to white.
4.
The phenotypic ratio is:
(A) 1:1 homozygous to heterozygous.
(B) 3:1 heterozygous to homozygous.
(C) All heterozygous.
(D) All black.
(E) 1:1 black to white.
5.
In lab, you cross a blue-flowered plant variety with a yellow-flowered variety. You get 50% blue plants in the F1
generation. The blue-flowered plants were probably :
(A) homozygous dominant.
(B) homozygous recessive.
(C) heterozygous.
(D) sterile.
(E) it is impossible to confirm the parental genotype from the information given.
6.
You could confirm the genotypes in the above question by:
(A) performing a test-cross.
(B) self-fertilizing the F1 generation.
(C) crossing with a homozygous recessive plant.
(D) crossing with a homozygous dominant plant.
(E) A and C.
7.
Which of the following represents an individual that is homozygous for a recessive trait?
(A) AA.
(B) aa.
(C) Aa.
(D) AA and Aa.
(E) none of the above.
The following statement is for questions 8 to 10.
7
Two common coat colors for Shetland sheepdogs are sable and tricolor. Sable dogs (BB) are blonde and tricolored dogs (bb) are
black with a white collar and tan eyebrows. A dog that is heterozygous for coat color (Bb) is called a tri-factored sable and is
darker than a dog that is homozygous for sable (BB).
8.
A sable dog is crossed with a tricolored dog. What are the possible colors of the progeny?
(A) All tri-factored sable.
(B) All sable.
(C) All blonde.
(D) Both sable and tricolored.
(E) both tricolored and tri-factored.
9.
One of the progeny from question 44 is crossed with a tricolored dog. What coat colors are possible?
(A) All tri-factored sable.
(B) All sable.
(C) All blonde.
(D) Both sable and tricolored.
(E) both tricolored and tri-factored.
10.
A tri-factored sheltie has a litter of puppies. Two of the pups are tri-factored and 2 of the pups are sable. What color was
the father of the puppies?
(A) sable.
(B) tricolored.
(C) tri-factored.
(D) either A or B.
(E) either B or C.
11.
A cross between 2 individuals results in a 9:3:3:1 ratio of phenotypes. This is a ___________ cross.
(A) monohybrid
(B) dihybrid
(C) testcross
(D) punnett square
(E) none of the above
12.
The degree of earlobe attachment in humans is inherited as a simple dominant of free earlobes, or as a recessive allele of
attached earlobes. What is the probability that a woman with attached earlobes will have children with attached earlobes
if their father has free earlobes, but his father had attached earlobes?
(A) 1 in 2.
(B) 3 in 4.
(C) 100%.
(D) 1 in 10.
(E) none.
13.
Alleles are:
(A) homologous chromosomes.
(B) alternate forms of a gene.
(C) the appearance of an individual.
(D) the genotype of an individual.
(E) heterozygous individuals.
14.
Mendel’s second law is also known as:
(A) the law of segregation.
(B) the law of independent assortment.
(C) the first law of thermodynamics.
(D) meiosis.
(E) increasing entropy.
8
15.
The law of segregation refers to:
(A) separation of gametes.
(B) separation of homologous chromosomes in anaphase I.
(C) separation of homologous chromosomes in metaphase I.
(D) independent assortment in metaphase I.
(E) independent assortment in metaphase II.
16.
A Rh-negative mother pregnant with her second Rh-positive child may cause the red blood cells of the fetus to clump.
This is because the mother has:
(A) anti-A antibodies
(B) anti-B antibodies
(C) anti-Rh antibodies
(D) an allergic response to pregnancy
(E) HIV
17.
A woman with blood type A has a child with a man that is blood type B. What is the possible genotype of the child?
(A) A.
(B) B.
(C) AB.
(D) O.
(E) all of the above.
The following statement is for questions 18 and 19.
Red-green colorblindness is a sex-linked trait. A woman with normal color vision whose father was colorblind has children with a
colorblind man.
18.
What is the probability that one of their children will be a colorblind female?
(A) 25%.
(B) 50%.
(C) 75%.
(D) 100%.
(E) 0.
19.
What is the probability that a son will have normal color vision?
(A) 25%.
(B) 50%.
(C) 75%.
(D) 100%.
(E) 0.
20.
In snapdragons, pink-flowered plants are produced when red-flowered plants are crossed with white-flowered plants. This
type of inheritance can be described as:
(A) a completely dominant trait.
(B) simple dominant and recessive traits.
(C) codominance.
(D) incomplete dominance.
(E) A and B.
21.
Autosomes are:
(A) homologous chromosomes that have failed to separate in meiosis.
(B) chromosomes that are not involved in sex determination.
(C) sex chromosomes.
(D) sex-linked genes.
(E) recessively inherited disorders.
9
22.
The sex of a child is determined by:
(A) the female gamete.
(B) the male gamete.
(C) sex chromosomes.
(D) temperature.
(E) B and C.
23.
Which of the following is not true of sickle cell anemia?
(A) It is inherited as a recessive condition.
(B) Heterozygotes have the disease.
(C) It is caused by a polar amino acid being replaced by a non-polar amino acid.
(D) It is particularly prevalent amongst blacks.
(E) None of the above – they are all true.
24.
The type of hemophilia that occurs in the descendents of Queen Victoria:
(A) is sex-linked.
(B) is carried on the Y chromosome.
(C) causes the red blood cells to sickle.
(D) causes the red blood cells to clump.
(E) all of the above.
25.
Amniocentesis:
(A) allows prenatal diagnosis of many genetic disorders.
(B) allows a fetus to be imaged.
(C) is a procedure involved in genetic engineering.
(D) is a term for the binding of the codon to the anticodon.
(E) none of the above.
26.
The maintenance of the sickle cell allele in areas with a high incidence of falciparum malaria is an example of:
(A) adaptation.
(B) disruptive selection.
(C) stabilizing selection.
(D) directional selection.
(E) A and C.
27.
The allele for Huntington’s disease persists because:
(A) it is recessive.
(B) it increases resistance to malaria.
(C) symptoms of the disease do not become evident until middle age.
(D) females are only carriers, they do not have the disease.
(E) A and B.
28.
Humans with an abnormal chromosome number are said to be:
(A) monosomic.
(B) trisomic.
(C) haploid.
(D) aneuploid.
(E) asexual.
29.
The organization of the chromosomes of an individual can be shown by a(n):
(A) aneuploid.
(B) ultrasound.
(C) pleiotrope.
(D) purine.
(E) karyotype.
10
29.
A person with Klinefelter syndrome is a:
(A) sterile female, short in stature, with a webbed neck.
(B) sterile, but otherwise normal female.
(C) sterile male with female characteristics.
(D) dead male.
(E) a male that is fertile but has an extra Y chromosome.
30.
The genotype of a person with Turner syndrome is:
(A) OY.
(B) OX.
(C) XXX.
(D) XXY.
(E) XYY.
11