Thursday, September 25, 2014

SOAL GENETIKA EXPERT GRADE 1

1) When Thomas Hunt Morgan crossed his red-eyed F₁ generation flies to each other, the F₂ generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result?
A) The gene involved is on the Y chromosome.
B) The gene involved is on the X chromosome.
C) The gene involved is on an autosome, but only in males.
D) Other male-specific factors influence eye color in flies.
E) Other female-specific factors influence eye color in flies.
Answer: B

2) Sturtevant provided genetic evidence for the existence of four pairs of chromosomes in Drosophila in which of these ways?
A) There are four major functional classes of genes in Drosophila.
B) Drosophila genes cluster into four distinct groups of linked genes.
C) The overall number of genes in Drosophila is a multiple of four.
D) The entire Drosophila genome has approximately 400 map units.
E) Drosophila genes have, on average, four different alleles.
Answer: B

3) Which of the following is the meaning of the chromosome theory of inheritance as expressed in the early 20th century?
A) Individuals inherit particular chromosomes attached to genes.
B) Mendelian genes are at specific loci on the chromosome and in turn segregate during meiosis.
C) Homologous chromosomes give rise to some genes and crossover chromosomes to other genes.
D) No more than a single pair of chromosomes can be found in a healthy normal cell.
E) Natural selection acts on certain chromosome arrays rather than on genes.
Answer: B

4) Thomas Hunt Morgan's choice of Drosophila melanogaster has been proven to be useful even today. Which of the following has/have continued to make it a most useful species?
I. its four pairs of chromosomes
II. a very large number of visible as well as biochemically mutant phenotypes
III. easy and inexpensive maintenance
IV. short generation time and large number of offspring
A) I and IV only
B) II and III only
C) I, II, and III only
D) II, III, and IV only
E) I, II, III, IV, and V
Answer: E

5) A woman is found to have 47 chromosomes, including three X chromosomes. Which of the following describes her expected phenotype?
A) masculine characteristics such as facial hair
B) enlarged genital structures
C) excessive emotional instability
D) normal female
E) sterile female
Answer: D

6) Males are more often affected by sex-linked traits than females because
A) male hormones such as testosterone often alter the effects of mutations on the X chromosome.
B) female hormones such as estrogen often compensate for the effects of mutations on the X chromosome.
C) X chromosomes in males generally have more mutations than X chromosomes in females.
D) males are hemizygous for the X chromosome.
E) mutations on the Y chromosome often worsen the effects of X-linked mutations.
Answer: D

7) SRY is best described in which of the following ways?
A) a gene present on the X chromosome that triggers female development
B) an autosomal gene that is required for the expression of genes on the Y chromosome
C) a gene region present on the Y chromosome that triggers male development
D) an autosomal gene that is required for the expression of genes on the X chromosome
E) a gene required for development, and males or females lacking the gene do not survive past early childhood
Answer: C

8) In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male?
A) tortoiseshell females; tortoiseshell males
B) black females; orange males
C) orange females; orange males
D) tortoiseshell females; black males
E) orange females; black males
Answer: D

9) Red-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son. What are the genotypes of the parents?
A) XcXc and XcY
B) XcXc and XCY
C) XCXC and XcY
D) XCXC and XCY
E) XCXc and XCY
Answer: E

10) Cinnabar eyes is a sex-linked recessive characteristic in fruit flies. If a female having cinnabar eyes is crossed with a wild-type male, what percentage of the F₁ males will have cinnabar eyes?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: E

11) Calico cats are female because
A) the males die during embryonic development.
B) a male inherits only one of the two X-linked genes controlling hair color.
C) the Y chromosome has a gene blocking orange coloration.
D) only females can have Barr bodies.
E) multiple crossovers on the Y chromosome prevent orange pigment production.
Answer: B

12) In birds, sex is determined by a ZW chromosome scheme. Males are ZZ and females are ZW. A recessive lethal allele that causes death of the embryo is sometimes present on the Z chromosome in pigeons. What would be the sex ratio in the offspring of a cross between a male that is heterozygous for the lethal allele and a normal female?
A) 2:1 male to female
B) 1:2 male to female
C) 1:1 male to female
D) 4:3 male to female
E) 3:1 male to female
Answer: A

13) Sex determination in mammals is due to the SRY region of the Y chromosome. An abnormality of this region could allow which of the following to have a male phenotype?
A) Turner syndrome, 45, X
B) translocation of SRY to an autosome of a 46, XX individual
C) a person with an extra X chromosome
D) a person with one normal and one shortened (deleted) X
E) Down syndrome, 46, XX
Answer: B

14) In humans, clear gender differentiation occurs, not at fertilization, but after the second month of gestation. What is the first event of this differentiation?
A) formation of testosterone in male embryos
B) formation of estrogens in female embryos
C) anatomical differentiation of a penis in male embryos
D) activation of SRY in male embryos and masculinization of the gonads
E) activation of SRY in females and feminization of the gonads
Answer: D

15) Duchenne muscular dystrophy (DMD) is caused by a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin, a muscle protein. They rarely live past their 20s. How likely is it for a woman to have this condition?
A) Women can never have this condition.
B) One-half of the daughters of an affected man could have this condition.
C) One-fourth of the children of an affected father and a carrier mother could have this condition.
D) Very rarely would a woman have this condition; the condition would be due to a chromosome error.
E) Only if a woman is XXX could she have this condition.
Answer: D

16) Women (and all female mammals) have one active X chromosome per cell instead of two. What causes this?
A) modification of the XIST gene so that it is active only on one X chromosome, which then becomes inactive
B) activation of the Barr gene on one of the two X chromosomes that then inactivates
C) crossover between the XIST gene on one X chromosome and a related gene on an autosome
D) inactivation of the XIST gene on the X chromosome derived from the male parent
E) the removal of methyl (CH3) groups from the X chromosome that will remain active
Answer: A

17) Which of the following statements is true of linkage?
A) The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them.
B) The observed frequency of recombination of two genes that are far apart from each other has a maximum value of 100%.
C) All of the traits that Mendel studied–seed color, pod shape, flower color, and others–are due to genes linked on the same chromosome.
D) Linked genes are found on different chromosomes.
E) Crossing over occurs during prophase II of meiosis.
Answer: A

18) How would one explain a testcross involving F₁ dihybrid flies in which more parental-type offspring than recombinant-type offspring are produced?
A) The two genes are closely linked on the same chromosome.
B) The two genes are linked but on different chromosomes.
C) Recombination did not occur in the cell during meiosis.
D) The testcross was improperly performed.
E) Both of the characters are controlled by more than one gene.
Answer: A

19) What does a frequency of recombination of 50% indicate?
A) The two genes are likely to be located on different chromosomes.
B) All of the offspring have combinations of traits that match one of the two parents.
C) The genes are located on sex chromosomes.
D) Abnormal meiosis has occurred.
E) Independent assortment is hindered.
Answer: A

20) What is the reason that linked genes are inherited together?
A) They are located close together on the same chromosome.
B) The number of genes in a cell is greater than the number of chromosomes.
C) Chromosomes are unbreakable.
D) Alleles are paired together during meiosis.
E) Genes align that way during metaphase I of meiosis.
Answer: A

21) Three genes at three loci are being mapped in a particular species. Each has two phenotypes, one of which is markedly different from the wild type. The unusual allele of the first gene is inherited with either of the others about 50% of the time. However, the unusual alleles of the other two genes are inherited together 14.4% of the time. Which of the following describes what is happening?
A) The genes are showing independent assortment.
B) The three genes are linked.
C) The first gene is linked but the other two are not.
D) The first gene is assorting independently from the other two that are linked.
E) The first gene is located 14.4 units apart from the other two.
Answer: D

22) The centimorgan (cM) is a unit named in honor of Thomas Hunt Morgan. To what is it equal?
A) the physical distance between two linked genes
B) 1% frequency of recombination between two genes
C) 1 nanometer of distance between two genes
D) the distance between a pair of homologous chromosomes
E) the recombination frequency between two genes assorting independently
Answer: B

23) Recombination between linked genes comes about for what reason?
A) Mutation on one homolog is different from that on the other homolog.
B) Independent assortment sometimes fails because Mendel had not calculated appropriately.
C) When genes are linked they always "travel" together at anaphase.
D) Crossovers between these genes result in chromosomal exchange.
E) Nonrecombinant chromosomes break and then re-join with one another.
Answer: D

24) Why does recombination between linked genes continue to occur?
A) Recombination is a requirement for independent assortment.
B) Recombination must occur or genes will not assort independently.
C) New allele combinations are acted upon by natural selection.
D) The forces on the cell during meiosis II always result in recombination.
E) Without recombination there would be an insufficient number of gametes.
Answer: C

25) Map units on a linkage map cannot be relied upon to calculate physical distances on a chromosome for which of the following reasons?
A) The frequency of crossing over varies along the length of the chromosome.
B) The relationship between recombination frequency and map units is different in every individual.
C) Physical distances between genes change during the course of the cell cycle.
D) The gene order on the chromosomes is slightly different in every individual.
E) Linkage map distances are identical between males and females.
Answer: A

26) Which of the following two genes are closest on a genetic map of Drosophila?
A) b and vg
B) vg and cn
C) rb and cn
D) cn and b
E) b and rb
Answer: E

27) If nondisjunction occurs in meiosis II during gametogenesis, what will be the result at the completion of meiosis?
A) All the gametes will be diploid.
B) Half of the gametes will be n + 1, and half will be n - 1.
C) 1/4 of the gametes will be n + 1, 1/4 will be n - 1, and 1/2 will be n.
D) There will be three extra gametes.
E) Two of the four gametes will be haploid, and two will be diploid.
Answer: C

28) One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. What is this alteration called?
A) deletion
B) transversion
C) inversion
D) translocation
E) duplication
Answer: D

29) A nonreciprocal crossover causes which of the following products?
A) deletion only
B) duplication only
C) nondisjunction
D) deletion and duplication
E) duplication and nondisjunction
Answer: D

30) In humans, male-pattern baldness is controlled by an autosomal gene that occurs in two allelic forms. Allele Hn determines nonbaldness, and allele Hb determines pattern baldness. In males, because of the presence of testosterone, allele Hb is dominant over Hn. If a man and woman both with genotype HnHb have a son, what is the chance that he will eventually be bald?
A) 0%
B) 25%
C) 33%
D) 50%
E) 75%
Answer: E

31) Of the following human aneuploidies, which is the one that generally has the most severe impact on the health of the individual?
A) 47, +21
B) 47, XXY
C) 47, XXX
D) 47, XYY
E) 45, X
Answer: A

32) A phenotypically normal prospective couple seeks genetic counseling because the man knows that he has a translocation of a portion of his chromosome 4 that has been exchanged with a portion of his chromosome 12. Although he is normal because his translocation is balanced, he and his wife want to know the probability that his sperm will be abnormal. What is your prognosis regarding his sperm?
A) 1/4 will be normal, 1/4 will have the translocation, and 1/2 will have duplications and deletions.
B) All will carry the same translocation as the father.
C) None will carry the translocation since abnormal sperm will die.
D) His sperm will be sterile and the couple might consider adoption.
E) 1/2 will be normal and the rest will have the father's translocation.
Answer: A

33) Abnormal chromosomes are frequently found in malignant tumors. Errors such as translocations may place a gene in close proximity to different control regions. Which of the following might then occur to make the cancer worse?
A) an increase in nondisjunction
B) expression of inappropriate gene products
C) a decrease in mitotic frequency
D) death of the cancer cells in the tumor
E) sensitivity of the immune system
Answer: B

34) An inversion in a human chromosome often results in no demonstrable phenotypic effect in the individual. What else may occur?
A) There may be deletions later in life.
B) Some abnormal gametes may be formed.
C) There is an increased frequency of mutation.
D) All inverted chromosomes are deleted.
E) The individual is more likely to get cancer.
Answer: B

35) What is the source of the extra chromosome 21 in an individual with Down syndrome?
A) nondisjunction in the mother only
B) nondisjunction in the father only
C) duplication of the chromosome
D) nondisjunction or translocation in either parent
E) It is impossible to detect with current technology.
Answer: D

36) Down syndrome has a frequency in the U.S. population of ~1/700 live births. In which of the following groups would you expect this frequency to be significantly higher?
A) people in Latin or South America
B) the Inuit and other peoples in very cold habitats
C) people living in equatorial areas of the world
D) very small population groups
E) No groups have such higher frequency.
Answer: E

37) A couple has a child with Down syndrome. The mother is 39 years old at the time of delivery. Which of the following is the most probable cause of the child's condition?
A) The woman inherited this tendency from her parents.
B) One member of the couple carried a translocation.
C) One member of the couple underwent nondisjunction in somatic cell production.
D) One member of the couple underwent nondisjunction in gamete production.
E) The mother had a chromosomal duplication.
Answer: D

38) In 1956 Tijo and Levan first successfully counted human chromosomes. What is the reason it took so many years to do so?
A) Watson and Crick's structure of DNA was not done until 1953.
B) Chromosomes were piled up on top of one another in the nucleus.
C) Chromosomes were not distinguishable during interphase.
D) A method had not yet been devised to halt mitosis at metaphase.
E) Chromosomes were piled up on top of one another in the nucleus, chromosomes were not distinguishable during interphase, and a method had not yet been devised to halt mitosis at metaphase.
Answer: E

39) At which phase(s) is it preferable to obtain chromosomes to prepare a karyotype?
A) early prophase
B) late telophase
C) anaphase
D) late anaphase or early telophase
E) late prophase or metaphase
Answer: E

40) What is a syndrome?
A) a characteristic facial appearance
B) a group of traits, all of which must be present if an aneuploidy is to be diagnosed
C) a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation
D) a characteristic trait usually given the discoverer's name
E) a characteristic that only appears in conjunction with one specific aneuploidy
Answer: C

41) Which of the following is known as a Philadelphia chromosome?
A) a human chromosome 22 that has had a specific translocation
B) a human chromosome 9 that is found only in one type of cancer
C) an animal chromosome found primarily in the mid-Atlantic area of the United States
D) an imprinted chromosome that always comes from the mother
E) a chromosome found not in the nucleus but in mitochondria
Answer: A

42) At what point in cell division is a chromosome lost so that, after fertilization with a normal gamete, the result is an embryo with 45, X?
I. an error in anaphase I
II. an error in anaphase II
III. an error of the first postfertilization mitosis
IV. an error in pairing
A) I or II only
B) II or IV only
C) III or IV only
D) I, II, or III only
E) I, II, III, or IV
Answer: E

43) Which of the following is true of aneuploidies in general?
A) A monosomy is more frequent than a trisomy.
B) 45 X is the only known human live-born monosomy.
C) Some human aneuploidies have selective advantage in some environments.
D) Of all human aneuploidies, only Down syndrome is associated with mental retardation.
E) An aneuploidy resulting in the deletion of a chromosome segment is less serious than a duplication.
Answer: B

44) A gene is considered to be non-Mendelian in its inheritance pattern if it seems to "violate" Mendel's laws. Which of the following would be considered Mendelian?
A) a gene whose expression varies depending on the gender of the transmitting parent
B) a gene derived solely from maternal inheritance
C) a gene transmitted via the cytoplasm or cytoplasmic structures
D) a gene transmitted to males from the maternal line and from fathers to daughters
E) a gene transmitted by a virus to egg-producing cells
Answer: D

45) Genomic imprinting is generally due to the addition of methyl (–CH3) groups to C nucleotides in order to silence a given gene. If this depends on the sex of the parent who transmits the gene, which of the following must be true?
A) Methylation of C is permanent in a gene.
B) Genes required for early development stages must not be imprinted.
C) Methylation of this kind must occur more in males than in females.
D) Methylation must be reversible in ovarian and testicular cells.
E) The imprints are transmitted only to gamete-producing cells.
Answer: D

46) Correns described that the inheritance of variegated color on the leaves of certain plants was determined by the maternal parent only. What phenomenon does this describe?
A) mitochondrial inheritance
B) chloroplast inheritance
C) genomic imprinting
D) infectious inheritance
E) sex-linkage
Answer: B

47) Mitochondrial DNA is primarily involved in coding for proteins needed for electron transport. Therefore, in which body systems would you expect most mitochondrial gene mutations to be exhibited?
A) the immune system and the blood
B) the excretory and respiratory systems
C) the skin and senses
D) the nervous and muscular systems
E) the circulation system
Answer: D

48) A certain kind of snail can have a right-handed direction of shell coiling (D) or left-handed coiling (d). If direction of coiling is due to a protein deposited by the mother in the egg cytoplasm, then a Dd egg-producing snail and a dd sperm-producing snail will have offspring of which genotype(s) and phenotype(s)?
A) 1/2 Dd:1/2 dd; all right coiling
B) all Dd; all right coiling
C) 1/2 Dd:1/2 dd; half right and half left coiling
D) all Dd; all left coiling
E) all Dd; half right and half left coiling
Answer: A

49) Which of the following produces a Mendelian pattern of inheritance?
A) genomic imprinting
B) a mitochondrial gene mutation
C) a chloroplast gene mutation
D) viral genomes that inhabit egg cytoplasm
E) a trait acted upon by many genes
Answer: E

50) Suppose that a gene on human chromosome 18 can be imprinted in a given pattern in a female parent but not in a male parent. A couple in whom each maternal meiosis is followed by imprinting of this gene have children. What can we expect as a likely outcome?
A) All sons but no daughters will bear their mother's imprinting pattern.
B) All daughters but no sons will bear their mother's imprinting pattern.
C) All sons and daughters will have a 50% chance of receiving the mother's imprinting pattern.
D) All the children will bear their mother's imprinting pattern but only daughters will then pass it down.
E) Each of the children will imprint a different chromosome.
Answer: D

51. This a map of four genes on a chromosome (See Image)

Between which two genes would you expect the highest frequency of recombination?
A) A and W
B) W and E
C) E and G
D) A and E
E) A and G
Answer: E

52) In a series of mapping experiments, the recombination frequencies for four different linked genes of Drosophila were determined as shown in Figure
What is the order of these genes on a chromosome map?
A) rb-cn-vg-b
B) vg-b-rb-cn
C) cn-rb-b-vg
D) b-rb-cn-vg
E) vg-cn-b-rb
Answer: D

53) The pedigree in Figure shows the transmission of a trait in a particular family.



Based on this pattern of transmission, the trait is most likely
A) mitochondrial.
B) autosomal recessive.
C) sex-linked dominant.
D) sex-linked recessive.
E) autosomal dominant.
Answer: A

54. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive.

How many of their daughters might be expected to be color-blind dwarfs?
A) all
B) none
C) half
D) one out of four
E) three out of four
Answer: B

55. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive.
What proportion of their sons would be color-blind and of normal height?
A) none
B) half
C) one out of four
D) three out of four
E) all
Answer: B

56. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive.
They have a daughter who is a dwarf with normal color vision. What is the probability that she is heterozygous for both genes?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: E

57. A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows:


A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type):
Which of the following are the phenotypes of the parents in this cross?
A) 2 and 5
B) 1 and 6
C) 4 and 8
D) 3 and 7
E) 1 and 2
Answer: C

58. A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows:
A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type): 


In which progeny phenotypes has there been recombination between genes A and B?
A) 1, 2, 5, and 6
B) 1, 3, 6, and 7
C) 2, 4, 5, and 8
D) 2, 3, 5, and 7
E) in all 8 of them
Answer: A

59. A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows:

A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type):


If recombination is equal to distance in centimorgans (cM), what is the approximate distance between genes A and B?
A) 1.5 cM
B) 3 cM
C) 6 cM
D) 15 cM
E) 30 cM
Answer: B

60. A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows:

A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type): 


What is the greatest benefit of having used a testcross for this experiment?
A) The homozygous recessive parents are obvious to the naked eye.
B) The homozygous parents are the only ones whose crossovers make a difference.
C) Progeny can be scored by their phenotypes alone.
D) All of the progeny will be heterozygous.
E) The homozygous recessive parents will be unable to cross over.
Answer: C

61. A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows:

A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type): 


The greatest distance among the three genes is between a and c. What does this mean?
A) Gene a is closest to b.
B) Genes are in the order: a–b–c.
C) Gene a is not recombining with c.
D) Gene a is between b and c.

E) Distance a–b is equal to distance a–c.

SOAL BIOLOGI GENETIKA SEDERHANA


1) What do we mean when we use the terms monohybrid cross and dihybrid cross?
A) A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents.
B) A monohybrid cross produces a single progeny, whereas a dihybrid cross produces two progeny.
C) A dihybrid cross involves organisms that are heterozygous for two characters and a monohybrid cross involves only one.
D) A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations.
E) A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.
Answer: C

2) Why did the F₁ offspring of Mendel's classic pea cross always look like one of the two parental varieties?
A) No genes interacted to produce the parental phenotype.
B) Each allele affected phenotypic expression.
C) The traits blended together during fertilization.
D) One phenotype was completely dominant over another.
E) Different genes interacted to produce the parental phenotype.
Answer: D

3) What was the most significant conclusion that Gregor Mendel drew from his experiments with pea plants?

A) There is considerable genetic variation in garden peas.
B) Traits are inherited in discrete units, and are not the results of "blending."
C) Recessive genes occur more frequently in the F₁ generation than do dominant ones.
D) Genes are composed of DNA.
E) An organism that is homozygous for many recessive traits is at a disadvantage.
Answer: B

4) How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?
A) 4
B) 8
C) 16
D) 32
E) 64
Answer: B

5) The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason?
A) segregation of maternal and paternal alleles
B) recurrent mutations forming new alleles
C) crossing over during prophase I
D) different possible alignments of chromosomes
E) the tendency for dominant alleles to segregate together
Answer: D

6) Why did Mendel continue some of his experiments to the F₂ or F₃ generation?
A) to obtain a larger number of offspring on which to base statistics
B) to observe whether or not a recessive trait would reappear
C) to observe whether or not the dominant trait would reappear
D) to distinguish which alleles were segregating
E) to be able to describe the frequency of recombination
Answer: B

7) Which of the following differentiates between independent assortment and segregation?
A) The law of independent assortment requires describing two or more genes relative to one another.
B) The law of segregation requires describing two or more genes relative to one another.
C) The law of segregation requires having two or more generations to describe.
D) The law of independent assortment is accounted for by observations of prophase I.
E) The law of segregation is accounted for by anaphase of mitosis.
Answer: A

8) Two plants are crossed, resulting in offspring with a 3:1 ratio for a particular trait. What does this suggest?
A) that the parents were true-breeding for contrasting traits
B) that the trait shows incomplete dominance
C) that a blending of traits has occurred
D) that the parents were both heterozygous for a single trait
E) that each offspring has the same alleles for each of two traits
Answer: D

9) A sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which of the following genotypes is possible in a gamete from this organism?
A) tt
B) Hh
C) HhTt
D) T
E) HT
Answer: E

10) When crossing an organism that is homozygous recessive for a single trait with a heterozygote, what is the chance of producing an offspring with the homozygous recessive phenotype?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: C

11) Mendel accounted for the observation that traits which had disappeared in the F₁ generation reappeared in the F₂ generation by proposing that
A) new mutations were frequently generated in the F₂ progeny, "reinventing" traits that had been lost in the F₁.
B) the mechanism controlling the appearance of traits was different between the F₁ and the F₂ plants.
C) traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F₁.
D) the traits were lost in the F₁ due to dominance of the parental traits.
E) members of the F₁ generation had only one allele for each trait, but members of the F₂ had two alleles for each trait.
Answer: C

12) The fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent assortment most probably indicates which of the following?
A) None of the traits obeyed the law of segregation.
B) The diploid number of chromosomes in the pea plants was 7.
C) All of the genes controlling the traits were located on the same chromosome.
D) All of the genes controlling the traits behaved as if they were on different chromosomes.
E) The formation of gametes in plants occurs by mitosis only.
Answer: D

13) Mendel was able to draw his ideas of segregation and independent assortment because of the influence of which of the following?
A) His reading and discussion of Darwin's Origin of Species.
B) The understanding of particulate inheritance he learned from renowned scientists of his time.
C) His discussions of heredity with his colleagues at major universities.
D) His experiments with the breeding of plants such as peas and fuchsia.
E) His reading of the scientific literature current in the field.
Answer: E

14) Mendel's observation of the segregation of alleles in gamete formation has its basis in which of the following phases of cell division?
A) prophase I of meiosis
B) anaphase II of meiosis
C) metaphase I of meiosis
D) anaphase I of meiosis
E) anaphase of mitosis
Answer: D

15) Mendel's second law of independent assortment has its basis in which of the following events of meiosis I?
A) synapsis of homologous chromosomes
B) crossing over
C) alignment of tetrads at the equator
D) separation of homologs at anaphase
E) separation of cells at telophase
Answer: C

16) Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of crosses BbTt × BBtt will be expected to have black fur and long tails?
A) 1/16
B) 3/16
C) 3/8
D) 1/2
E) 9/16
Answer: D

17) In certain plants, tall is dominant to short. If a heterozygous plant is crossed with a homozygous tall plant, what is the probability that the offspring will be short?
A) 1
B) 1/2
C) 1/4
D) 1/6
E) 0
Answer: E

18) In the cross AaBbCc × AaBbCc, what is the probability of producing the genotype AABBCC?
A) 1/4
B) 1/8
C) 1/16
D) 1/32
E) 1/64
Answer: E

19) Given the parents AABBCc × AabbCc, assume simple dominance for each trait and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent?
A) 1/4
B) 1/8
C) 3/4
D) 3/8
E) 1
Answer: C

20) Which of the following is the best statement of the use of the addition rule of probability?
A) the probability that two or more independent events will both occur
B) the probability that two or more independent events will both occur in the offspring of one set of parents
C) the probability that either one of two independent events will occur
D) the probability of producing two or more heterozygous offspring
E) the likelihood that a trait is due to two or more meiotic events
Answer: C

21) Which of the following calculations require that you utilize the addition rule?
A) Calculate the probability of black offspring from the cross AaBb × AaBb, when B is the symbol for black.
B) Calculate the probability of children with both cystic fibrosis and polydactyly when parents are each heterozygous for both genes.
C) Calculate the probability of each of four children having cystic fibrosis if the parents are both heterozygous.
D) Calculate the probability of a child having either sickle-cell anemia or cystic fibrosis if parents are each heterozygous for both.
E) Calculate the probability of purple flower color in a plot of 50 plants seeded from a self-fertilizing heterozygous parent plant.
Answer: D

22) In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white?

A) red × white
B) roan × roan
C) white × roan
D) red × roan
E) The answer cannot be determined from the information provided.
Answer: B

23) Which of the following describes the ability of a single gene to have multiple phenotypic effects?
A) incomplete dominance
B) multiple alleles
C) pleiotropy
D) epistasis
Answer: C

24) Cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections. Which of the following terms best describes this?
A) incomplete dominance
B) multiple alleles
C) pleiotropy
D) epistasis
E) codominance
Answer: C

25) Which of the following is an example of polygenic inheritance?
A) pink flowers in snapdragons
B) the ABO blood group in humans
C) Huntington's disease in humans
D) white and purple flower color in peas
E) skin pigmentation in humans
Answer: E

26) Hydrangea plants of the same genotype are planted in a large flower garden. Some of the plants produce blue flowers and others pink flowers. This can be best explained by which of the following?
A) the knowledge that multiple alleles are involved
B) the allele for blue hydrangea being completely dominant
C) the alleles being codominant
D) the fact that a mutation has occurred
E) environmental factors such as soil pH
Answer: E

27) Which of the following provides an example of epistasis?
A) Recessive genotypes for each of two genes (aabb) results in an albino corn snake.
B) The allele b17 produces a dominant phenotype, although b1 through b16 do not.
C) In rabbits and many other mammals, one genotype (cc) prevents any fur color from developing.
D) In Drosophila (fruit flies), white eyes can be due to an X-linked gene or to a combination of other genes.
E) In cacti, there are several genes for the type of spines.
Answer: C

28) Most genes have many more than two alleles. However, which of the following is also true?

A) At least one allele for a gene always produces a dominant phenotype.
B) Most of the alleles will never be found in a live-born organism.
C) All of the alleles but one will produce harmful effects if homozygous.
D) There may still be only two phenotypes for the trait.
E) More than two alleles in a genotype is considered lethal.
Answer: D

29) How could you best predict the maximum number of alleles for a single gene whose polypeptide product is known?
A) Search the population for all phenotypic variants of this polypeptide.
B) Count the number of amino acids in the polypeptide.
C) Mate all known genotypes and collect all possible offspring different from the parents.
D) Measure the rate of new mutations in the species and estimate the number since it first evolved.
E) Count the number of DNA nucleotides that are in the code for the polypeptides.
Answer: E

30) An ideal procedure for fetal testing in humans would have which of the following features?
A) the procedure that can be performed at the earliest time in the pregnancy
B) lowest risk procedure that would provide the most reliable information
C) the procedure that can test for the greatest number of traits at once
D) a procedure that provides a three-dimensional image of the fetus
E) a procedure that could test for the carrier status of the fetus
Answer: A

31) A scientist discovers a DNA-based test for one allele of a particular gene. This and only this allele, if homozygous, produces an effect that results in death at or about the time of birth. Of the following, which is the best use of this discovery?
A) Screen all newborns of an at-risk population.
B) Design a test for identifying heterozygous carriers of the allele.
C) Introduce a normal allele into deficient newborns.
D) Follow the segregation of the allele during meiosis.
E) Test school-age children for the disorder.
Answer: B

32) An obstetrician knows that one of her patients is a pregnant woman whose fetus is at risk for a serious disorder that is detectable biochemically in fetal cells. The obstetrician would most reasonably offer which of the following procedures to her patient?
A) CVS
B) ultrasound imaging
C) amniocentesis
D) blood transfusion
E) X-ray
Answer: C

33) The frequency of heterozygosity for the sickle-cell anemia allele is unusually high, presumably because this reduces the frequency of malaria. Such a relationship is related to which of the following?
A) Mendel's law of independent assortment
B) Mendel's law of segregation
C) Darwin's explanation of natural selection
D) Darwin's observations of competition
E) the malarial parasite changing the allele
Answer: C

34) Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize a particular amino acid. The amino acid is not otherwise produced by humans. Therefore, the most efficient and effective treatment is which of the following?
A) Feed them the substrate that can be metabolized into this amino acid.
B) Transfuse the patients with blood from unaffected donors.
C) Regulate the diet of the affected persons to severely limit the uptake of the amino acid.
D) Feed the patients the missing enzymes in a regular cycle, such as twice per week.
E) Feed the patients an excess of the missing product.
Answer: C

35) Hutchinson-Gilford progeria is an exceedingly rare human genetic disorder in which there is very early senility and death, usually from coronary artery disease, at an average age of approximately 13. Patients, who look very old even as children, do not live to reproduce. Which of the following represents the most likely assumption?
A) All cases must occur in relatives; therefore, there must be only one mutant allele.
B) Successive generations of a family will continue to have more and more cases over time.
C) The disorder may be due to mutation in a single protein-coding gene.
D) Each patient will have had at least one affected family member in a previous generation.
E) The disease is autosomal dominant.
Answer: C

36) One of two major forms of a human condition called neurofibromatosis (NF 1) is inherited as a dominant gene, although it may range from mildly to very severely expressed. If a young child is the first in her family to be diagnosed, which of the following is the best explanation?
A) The mother carries the gene but does not express it at all.
B) One of the parents has very mild expression of the gene.
C) The condition skipped a generation in the family.
D) The child has a different allele of the gene than the parents.
Answer: B

37. In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding dark-leaved plant is crossed with a light-leaved one, and the F₁ offspring is allowed to self-pollinate. The predicted outcome of the F₂ is diagrammed in the Punnett square shown in Figure Where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.


Which of the boxes marked 1-4 correspond to plants with dark leaves?
A) 1 only
B) 1 and 2
C) 2 and 3
D) 4 only
E) 1, 2, and 3
Answer: E


38. In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding dark-leaved plant is crossed with a light-leaved one, and the F₁ offspring is allowed to self-pollinate. The predicted outcome of the F₂ is diagrammed in the Punnett square shown in Figure Where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.

Which of the boxes correspond to plants with a heterozygous genotype?
A) 1
B) 1 and 2
C) 1, 2, and 3
D) 2 and 3
E) 2, 3, and 4
Answer: D

39. In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding dark-leaved plant is crossed with a light-leaved one, and the F₁ offspring is allowed to self-pollinate. The predicted outcome of the F₂ is diagrammed in the Punnett square shown in Figure Where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.

Which of the plants will be true-breeding?
A) 1 and 4 only
B) 2 and 3 only
C) 1, 2, 3, and 4
D) 1 only
E) 1 and 2 only
Answer: A

40. The following question refer to the pedigree chart in Figure


for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.
What is the genotype of individual II-5?
A) WW
B) Ww
C) ww
D) WW or ww
E) ww or Ww
Answer: C

41. The following question refer to the pedigree chart in Figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.


What is the likelihood that the progeny of IV-3 and IV-4 will have the trait?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: C

42. The following question refer to the pedigree chart in Figure  for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.


What is the probability that individual III-1 is Ww?
A) 3/4
B) 1/4
C) 2/4
D) 2/3
E) 1
Answer: E

43. Use the following pedigree (Figure ) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.



What is the genotype of the deceased individual in generation II?
A) homozygous for a gene for colon cancer
B) homozygous for both cancer alleles from his mother
C) heterozygous for a gene for colon cancer
D) affected by the same colon cancer environmental factor as his mother
E) carrier of all of the several known genes for colon cancer
Answer: C

44. Use the following pedigree (Figure ) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.

In each generation of this family after generation I, the age at diagnosis is significantly lower than would be found in nonfamilial (sporadic) cases of this cancer (~ 63 years). What is the most likely reason?
A) Members of this family know to be checked for colon cancer early in life.
B) Hereditary (or familial) cases of this cancer typically occur at earlier ages than do nonfamilial forms.
C) This is pure chance; it would not be expected if you were to look at a different family.
D) This cancer requires mutations in more than this one gene.
E) Affected members of this family are born with colon cancer, and it can be detected whenever they are first tested.
Answer: B

45. Use the following pedigree (Figure ) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.

From this pedigree, how does this trait seem to be inherited?
A) from mothers
B) as an autosomal recessive
C) as a result of epistasis
D) as an autosomal dominant
E) as an incomplete dominant
Answer: D

46. Use the following pedigree (Figure ) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.

The affected woman in generation IV is thinking about her future and asks her oncologist (cancer specialist) whether she can know whether any or all of her children will have a high risk of the same cancer. The doctor would be expected to advise which of the following?
I. genetic counseling
II. prenatal diagnosis when/if she becomes pregnant
III. testing to see whether she has the allele
IV. testing to see whether her future spouse or partner has the allele
A) I only
B) II only
C) I and II only
D) I, II, and III only
E) III and IV only
Answer: C

47. Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F₁ individuals have red, axial flowers. The genes for flower color and location assort independently.
 If 1,000 F₂ offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers?
A) 65
B) 190
C) 250
D) 565
E) 750
Answer: B

48. Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F₁ individuals have red, axial flowers. The genes for flower color and location assort independently.
Among the F₂ offspring, what is the probability of plants with white axial flowers?
A) 9/16
B) 1/16
C) 3/16
D) 1/8
E) 1/4
Answer: C

49. Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.
 These results indicate which of the following?
A) Brown is dominant to black.
B) Black is dominant to brown and to yellow.
C) Yellow is dominant to black.
D) There is incomplete dominance.
E) Epistasis is involved.
Answer: E

50. Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.

How many genes must be responsible for these coat colors in Labrador retrievers?
A) 1
B) 2
C) 3
D) 4
E) 5
Answer: B

51. Labrador retrievers are black, brown, or yellow. In a cross of a black female with a brown male, results can be either all black puppies, 1/2 black to 1/2 brown puppies, or 3/4 black to 1/4 yellow puppies.
In one type cross of black × black, the results were as follows:
9/16 black
4/16 yellow
3/16 brown
The genotype aabb must result in which of the following?
A) black
B) brown
C) yellow
D) a lethal result
E) white
Answer: C

52. Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant characteristic.
If true-breeding red long radishes are crossed with true-breeding white oval radishes, the F₁ will be expected to be which of the following?
A) red and long
B) red and oval
C) white and long
D) purple and long
E) purple and oval
Answer: D

53. Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant characteristic.
 In the F₂ generation of the above cross, which of the following phenotypic ratios would be expected?
A) 9:3:3:1
B) 9:4:3
C) 1:1:1:1
D) 1:1:1:1:1:1
E) 6:3:3:2:1:1
Answer: E

54. Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant characteristic.

The flower color trait in radishes is an example of which of the following?
A) a multiple allelic system
B) sex linkage
C) codominance
D) incomplete dominance
E) epistasis
Answer: D

55. Drosophila (fruit flies) usually have long wings (+) but mutations in two different genes can result in bent wings (bt) or vestigial wings (vg).
 If a homozygous bent wing fly is mated with a homozygous vestigial wing fly, which of the following offspring would you expect?
A) all +bt +vg heterozygotes
B) 1/2 bent and 1/2 vestigial flies
C) all homozygous + flies
D) 3/4 bent to 1/4 vestigial ratio
E) 1/2 bent and vestigial to 1/2 normal
Answer: A

56. Drosophila (fruit flies) usually have long wings (+) but mutations in two different genes can result in bent wings (bt) or vestigial wings (vg).
If flies that are heterozygous for both the bent wing gene and the vestigial wing gene are mated, what is the probability of offspring with bent wings only?
A) 1/8
B) 3/8
C) 1/4
D) 9/16
E) 3/16
Answer: E

57. Tallness (T) in snapdragons is dominant to dwarfness (t), while red (R) flower color is dominant to white (r). The heterozygous condition results in pink (Rr) flower color.
A dwarf, red snapdragon is crossed with a plant homozygous for tallness and white flowers. What are the genotype and phenotype of the F1 individuals?
A) ttRr—dwarf and pink
B) ttrr—dwarf and white
C) TtRr—tall and red
D) TtRr—tall and pink
E) TTRR—tall and red
Answer: D

58. Tallness (T) in snapdragons is dominant to dwarfness (t), while red (R) flower color is dominant to white (r). The heterozygous condition results in pink (Rr) flower color.
If snapdragons are heterozygous for height as well as for flower color, a mating between them will result in what ratio?
A) 9:3:3:1
B) 6:3:3:2:1:1
C) 1:2:1
D) 27:9:9:9:3:3:3:1
E) 9:4:3
Answer: B

59. Skin color in a certain species of fish is inherited via a single gene with four different alleles.
How many different types of gametes would be possible in this system?
A) 1
B) 2
C) 4
D) 8
E) 16
Answer: C

60. Skin color in a certain species of fish is inherited via a single gene with four different alleles.
One fish of this type has alleles 1 and 3 (S1S3) and its mate has alleles 2 and 4 (S2S4). If each allele confers a unit of color darkness such that S1 has one unit, S2 has two units, and so on, then what proportion of their offspring would be expected to have five units of color?
A) 1/4
B) 1/5
C) 1/8
D) 1/2
E) 0
Answer: D

61. Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.
The relationship between genes S and N is an example of
A) incomplete dominance.
B) epistasis.
C) complete dominance.
D) pleiotropy.
E) codominance.
Answer: B

62. Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.
A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce
A) all sharp-spined progeny.
B) 50% sharp-spined, 50% dull-spined progeny.
C) 25% sharp-spined, 50% dull-spined, 25% spineless progeny.
D) all spineless progeny.
E) It is impossible to determine the phenotypes of the progeny.
Answer: A

63. Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.
 If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the F2 would segregate in which of the following ratios?
A) 3 sharp-spined:1 spineless
B) 1 sharp-spined:2 dull-spined:1 spineless
C) 1 sharp-spined:1 dull-spined:1 spineless
D) 1 sharp-spined:1 dull-spined
E) 9 sharp-spined:3 dull-spined:4 spineless
Answer: E

64. Feather color in budgies is determined by two different genes, Y and B, one for pigment on the outside and one for the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white.
A blue budgie is crossed with a white budgie. Which of the following results is not possible?
A) green offspring only
B) yellow offspring only
C) blue offspring only
D) green and yellow offspring
E) a 9:3:3:1 ratio
Answer: D

65. Feather color in budgies is determined by two different genes, Y and B, one for pigment on the outside and one for the inside of the feather. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white.
Two blue budgies were crossed. Over the years, they produced 22 offspring, 5 of which were white. What are the most likely genotypes for the two blue budgies?
A) yyBB and yyBB
B) yyBB and yyBb
C) yyBb and yyBb
D) yyBB and yybb
E) yyBb and yybb
Answer: C

66. A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.
Which of the following is a possible partial genotype for the son?
A) IBIB
B) IBIA
C) ii
D) IBi
E) IAIA
Answer: D

67. A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.
Which of the following is a possible genotype for the mother?
A) IAIA
B) IBIB
C) ii
D) IAi
E) IAIB
Answer: D

68. A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.
Which of the following is a possible phenotype for the father?
A) A negative
B) O negative
C) B positive
D) AB negative
E) impossible to determine
Answer: C

69. A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.
If both children are of blood type M, which of the following is possible?
A) Each parent is either M or MN.
B) Each parent must be type M.
C) Both children are heterozygous for this gene.
D) Neither parent can have the N allele.
E) The MN blood group is recessive to the ABO blood group.
Answer: A

70. Humanoids on the newly explored planet Brin (in a hypothetical galaxy in ~50 years from the present) have a gene structure similar to our own, but many very different plants and animals.
One species of a small birdlike animal has an extremely variable tail length, which is a highly polymorphic trait. Geneticists have come to realize that there are eight separate genes for tail length per haploid genome, with each gene having two alleles. One allele for each gene (a, b, and so on) increases the length by 1 cm, whereas the other allele (a2, b2, and so on) increases it by 0.5 cm. One bird was analyzed and found to have the following genotype:
a1a1b2b2c1c2d1d2e2e2f1f2g1g1h1h2
What is the length of its tail?
A) 6 cm
B) 8 cm
C) 12 cm
D) 24 cm
E) 36 cm
Answer: C

71. Humanoids on the newly explored planet Brin (in a hypothetical galaxy in ~50 years from the present) have a gene structure similar to our own, but many very different plants and animals.
One species of green plant, with frondlike leaves, a spine-coated stem, and purple cup-shaped flowers, is found to be self-pollinating. Which of the following is true of this species?
A) The species must be haploid.
B) Its reproduction is asexual.
C) All members of the species have the same genotype.
D) Some of the seeds would have true-breeding traits.
E) All of its dominant traits are most frequent.
Answer: D

72. Humanoids on the newly explored planet Brin (in a hypothetical galaxy in ~50 years from the present) have a gene structure similar to our own, but many very different plants and animals.
72) If the environmental parameters, such as temperature, humidity, atmosphere, sunlight, and so on, are mostly Earthlike, which of the following do you expect of its types of leaves, stems, and flowers?
A) The genes for them would have originated on Earth.
B) Genes for these traits would have a common ancestor with those from Earth.
C) Such plants could be safely eaten by humans.
D) Genotypes for these traits would be identical to Earth plants with the same traits.
E) Phenotypes would be selected for or against by these environmental factors.
Answer: E

73. Humanoids on the newly explored planet Brin (in a hypothetical galaxy in ~50 years from the present) have a gene structure similar to our own, but many very different plants and animals.
Marfan syndrome in humans is caused by an abnormality of the connective tissue protein fibrillin. Patients are usually very tall and thin, with long spindly fingers, curvature of the spine, sometimes weakened arterial walls, and sometimes ocular problems, such as lens dislocation. Which of the following would you conclude about Marfan syndrome from this information?
A) It is recessive.
B) It is dominant.
C) It has a late age of onset (> 60).
D) It is pleiotropic.
E) It is epistatic
Answer : D.

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