Cytogenetics Practice Test
Review the following concepts if needed:
- Chromosome Structure for Genetics
- Variation in Chromosome Number for Genetics
- Variation in Chromosome Size for Genetics
- Variation in the Arrangement of Chromosome Segments for Genetics
- Variation in the Number of Chromosome Segments for Genetics
- Variation in Chromosome Morphology for Genetics
- Human Cytogenetics for Genetics
Cytogenetics Practice Test
For each of the following definitions, give the appropriate term and spell it correctly. Terms are single words unless indicated otherwise.
- A cell or organism containing three sets of chromosomes.
- A cell or organism produced by doubling the chromosome number of an interspecific hybrid.
- Any variation in chromosome number that does not involve whole sets of chromosomes.
- A cell or organism having a genomic formula 2n – 1.
- An adjective applicable to a giant chromosome consisting of hundreds of chromatid strands.
- Exchange of pieces between two nonhomologous chromosomes. (Two words.)
- Altered phenotypic expression of a gene as a consequence of movement from its normal location. (Two words.)
- Achromosomal aberration that, with the help of crossing over within the aberration, can lead to "bridge and fragment" formation. (Two words.)
- Phenotypic expression of a recessive gene as a consequence of loss of a chromosomal segment bearing the corresponding dominant allele.
- The arrangement of the somatic chromosome complement (karyotype) of a cell in groups of homologous pairs.
Choose the one best answer.
- A treatment often used to induce polyploidy experimentally in plants is (a) X-rays (b) gibberellic acid (c) colchicine (d) acridine dyes (e) azothioprene
- A mechanism that can cause a gene to move from one linkage group to another is (a) translocation (b) inversion (c) crossing over (d) duplication (e) dosage compensation
- If during synapsis a certain kind of abnormal chromosome is always forced to bulge away from its normal homologue, the abnormality is classified as (a) an inversion (b) a duplication (c) an isochromosome (d) a deficiency (e) none of the above
- If four chromosomes synapse into a cross-shaped configuration during meiotic prophase, the organism is heterozygous for a (a) pericentric inversion (b) deletion (c) translocation (d) paracentric inversion (e) none of the above
- A segment of chromosome may be protected from recombination by (a) an inversion (b) a translocation (c) balanced lethals (d) more than one of the above (e) all of the above
- A person with Klinefelter syndrome is considered to be (a) monosomic (b) triploid (c) trisomic (d) deletion heterozygote (e) none of the above
- Given a normal chromosome with segments labeled C123456 (C = centromere), a homologue containing an inversion including regions 3–5, and a single two-strand crossover between regions 4 and 5; then the acentric fragment present during first meiotic anaphase is (a) 63456 (b) 12344321 (c) 65521 (d) 654321 (e) none of the above
- Pseudodominance may be observed in heterozygotes for (a) a deletion (b) a duplication (c) a paracentric inversion (d) a reciprocal translocation (e) more than one of the above
- The most easily recognized characteristic of an inversion heterozygote in plants is (a) gigantism (b) semisterility (c) a cross-shaped chromosome configuration during meiosis (d) pseudodominance (e) none of the above
- If the garden pea has 14 chromosomes in its diploid complement, how many double trisomics could theoretically exist? (a) 6 (b) 9 (c) 16 (d) 21 (e) none of the above. (Hint: See formula (2.1).]
Variation in Chromosome Number Questions
- Abyssinian oat (Avena abyssinica) appears to be a tetraploid with 28 chromosomes. The common cultivated oat (Avena sativa) appears to be a hexaploid in this same series. How many chromosomes does the common oat possess?
- The European raspberry (Rubus idaeus) has 14 chromosomes. The dewberry (Rubus caesius) is a tetraploid with 28 chromosomes. Hybrids between these two species are sterile F1 individuals. Some unreduced gametes of the F1 are functional in backcrosses. Determine the chromosome number and level of ploidy for each of the following: (a) F1 (b) F1 back-crossed to R. idaeus (c) F1 backcrossed to R. caesius (d) chromosome doubling of F1 (R. maximus).
- There are 13 pairs of chromosomes in Asiatic cotton (Gossypium arboreum) and also 13 pairs in an American species G. thurberi. Interspecific crosses between arboreum and thurberi are sterile because of highly irregular chromosome pairing during meiosis. The American cultivated cotton (G. hirsutum) has 26 pairs of chromosomes. Crosses of arboreum × hirsutum or thurberi × hirsutum produce triploids with 13 bivalents (pairs of chromosomes) and 13 univalents (single unpaired chromosomes). How can this cytological information be used to interpret the evolution of G. hirsutum?
- If two alleles, A and a, exist at a locus, five genotypic combinations can be formed in an autotetraploid: quadruplex (AAAA), triplex (AAAa), duplex (AAaa), simplex (Aaaa), and nulliplex (aaaa). Assume that A exhibits xenia over a. For each of these five genotypes, determine the expected phenotypic ratio of A : a when, (a) the locus is tightly linked to its centromere (chromosomal assortment) and the genotype is selfed, (b) the locus is assorting chromosomally and the genotype is testcrossed, (c) the locus is far from its centromere so that chromatids assort independently and the genotype is selfed, (d) the locus assorts by chromatids and the genotype is testcrossed.
- The flinty endosperm character in maize is produced whenever two or all three of the alleles in this triploid tissue are F. In the presence of its alternative allele F' in double or triple dose, a floury endosperm is produced. White endosperm color is produced by a triple dose of a recessive allele y, its dominant allele Y exhibiting xenia and producing yellow endosperm. The loci of F and Y assort independently. (a) In crosses between parents of genotype FF'Yy, what phenotypic ratio is expected in the progeny seed? (b) Pollen from a plant of genotype FF'Yy is crossed onto a plant of genotype FFyy. Compare the phenotypic ratios produced by this cross with its reciprocal cross.
- The diploid number of an organism is 12. How many chromosomes would be expected in the following: (a) a monosomic (b) a trisomic (c) a tetrasomic (d) a double trisomic (e) a nullisomic (f) a monoploid (g) a triploid (h) an autotetraploid?
- Sugary endosperm of corn is regulated by a recessive gene s on chromosome IV and starchy endosperm by its dominant alllele S. Assuming n + 1 pollen grains are nonfunctional, predict the genotypic and phenotypic ratios of endosperm expected in the progeny from the cross of (a) diploid ss pollinated by trisomic-IV of genotype SSs, (b) diploid Ss pollinated by trisomic-IV of genotype SSs.
- A dominant gene w+ produces yellow flowers in a certain plant species and its recessive allele w produces white flowers. Plants trisomic for the chromosome bearing the color locus will produce n and n + 1 functional female gametes, but viable pollen has only the n number. Find the phenotypic ratio expected from each of the following crosses:
- Normal women possess two sex chromosomes (XX) and normal men have a single X chromosome plus a Y chromosome that carries male determiners. Rarely, a woman is found with marked abnormalities of primary and secondary sexual characteristics, and she has only one X chromosome (XO). The phenotypic expressions of this monosomic-X state is called Turner syndrome. Likewise, men are occasionally discovered with an XXY constitution that exhibits corresponding abnormalities and is called Klinefelter syndrome. Color blindness is a sex-linked recessive trait. (a)Ahusband and wife both have normal vision, but one of their children is a color-blind Turner girl. Diagram this cross, including the gametes that produced this child. (b) In another family the mother is color blind and the father has normal vision. Their child has Klinefelter's with normal vision. What gametes produced this child? (c) Suppose the same parents in part (b) produced a color-blind child with Klinefelter's. What gametes produced this child? (d) The normal diploid number for humans is 46. A trisomic condition for chromosome 21 results in Down syndrome. At least one case of Down-Klinefelter has been recorded. How many chromosomes would this individual be expected to possess?
Variation in Arrangement of Chromosome Segments Questions
- Colorless aleurone of corn kernels is a trait governed by a recessive gene c and is in the same linkage group (IX) with another recessive gene wx that governs waxy endosperm. In 1931, Creighton and McClintock found a plant with one normal IX chromosome, but its homologue had a knob on one end and a translocated piece from another chromosome on the other end. A dihybrid colored, starchy plant with the heteromorphic IX chromosome shown below was testcrossed to a colorless, waxy plant with normal chromosomes. The results of this experiment provided cytological proof that genetic crossing over involves an actual physical exchange between homologous chromosome segments. Diagram the results of this cross, showing all genotypes and phenotypes.
- Nipple-shaped tips on tomato fruit is the phenotypic expression of a recessive gene nt on chromosome V. A heterozygous plant (Nt/nt) that is also heterozygous for a reciprocal translocation involving chromosomes Vand VIII is testcrossed to a plant with normal chromosomes. The progeny were 48 normal fruit, fertile : 19 nipple fruit, fertile : 11 normal fruit, semisterile : 37 nipple fruit, semisterile. What is the genetic position of the locus of gene Nt with respect to the point of translocation?
- Given a pericentric inversion heterozygote with one chromosome in normal order (1 2 3 4.5 6 7 8) and the other in the inverted order (1 5.4 3 2 6 7 8), diagram the first anaphase figure after a four-strand double crossover occurs: one crossover involves the regions between 4 and the centromere (.); the other crossover occurs between the centromere and 5.
- Afour-strand double crossover occurs in an inversion heterozygote. The normal chromosome order is (.1 2 3 4 5 6 7 8); the inverted chromosome order is (.1 2 7 6 5 4 3 8).One crossover is between 1 and 2 and the other is between 5 and 6. Diagram and label the first anaphase figures.
- Achromosome with segments in the normal order is (.a b c d e f g h). An inversion heterozygote has the abnormal order (.a b f e d c g h). A three-strand double crossover occurs involving the regions between a and b and between d and e. Describe the first and second anaphase figures.
- A species of the fruit fly is differentiated into five races on the basis of differences in the banding patterns of one of its giant chromosomes. Eight regions of the chromosome are designated a–h. If each of these races is separated by a single overlapping inversion, devise a scheme to account for the evolution of the five races: (1) adghfcbe (2) fhgdacbe (3) fhcadgbe (4) fhgbcade (5) fadghcbe
Variation in the Number of Chromosome Segments Questions
- In higher animals, even very small deficiencies, when homozygous, are usually lethal. A recessive gene w in mice results in an abnormal gait called "waltzing." A waltzing male was crossed to several homozygous normal females. Among several hundred offspring, one was found to be a waltzer female. Presumably, a deficiency in the chromosome carrying the w+ allele caused the waltzing trait to appear as pseudodominant. The pseudodominant waltzer female was then crossed to a homozygous normal male and produced only normal offspring. (a) List two possible genotypes for the normal progeny from the above cross. (b) Suppose that two males, one of each genotype produced in part (a), were backcrossed to their pseudodominant mother and each produced 12 zygotes. Assuming that homozygosity for the deletion is lethal, calculate the expected combined number of waltzer and normal progeny.
Variation in Chromosome Morphology Questions
- Vermilion eye color in Drosophila is a sex-linked recessive condition; bar eye is a sex-linked dominant condition. An attached-X female with vermilion eyes, also having a Y chromosome (X^XY), is mated to a bar-eyed male. (a) Predict the phenotypic ratio that is expected in the F1 flies. (b) How much death loss is anticipated in the F1 generation? (Hint: see Problem 5.19) (c) What phenotypic ratio is expected in the F2?
- Two recessive sex-linked traits in Drosophila are garnet eye (g) and forked bristle (f). The attached-X chromosomes of females heterozygous for these genes are diagrammed below.
- Given the ring homozygote at the left (below), diagram the first anaphase figure when crossovers occur at position (a) A and B, (b) A and C, (c) A and D.
- Given the ring-rod heterozygote at the right (above), diagram the first anaphase figure when crossovers occur at positions (a) A and B, (b) A and C, (c) A and D.
A crossover between two chromatids attached to the same centromere is called a reciprocal exchange; a crossover between two chromatids attached to different centromeres is a non-reciprocal exchange. Approximately 7% of the daughters from these attached-X females were + + /f g, 7%were f + / + +, 7% were f g / + g and the remainder were f + / + g. (a)Which of the single exchanges (A, B, C, or D in the diagram) could produce the daughters (1) + + / f g and f + / + g, (2) f + / + + and f g / + g? (b) Are chromatids attached to the same centromere more likely to be involved in an exchange than chromatids attached to different centromeres? (c) Does the fact that neither homozygous wild-type nor garnet-forked progeny were found shed any light on the number of chromatids that undergo exchange at any one locus?
Human Cytogenetics Questions
- Meiotic nondisjunction of the sex chromosomes in either parent can produce a child with Klinefelter syndrome (XXY) or Turner syndrome (XO). Color blindness is due to a sex-linked recessive gene. (a) If a color-blind woman and man with normal vision produce a color-blind child, with Kleinfelter's in which parent did the nondisjunctional event occur? (b) If a heterozygous woman with normal vision and a man with normal vision produce a color-blind child, with Kleinfelter's how can this be explained?
- Explain what type of abnormal sperm unites with a normal egg to produce an XYY offspring. Specifically, how does such an abnormal gamete arise?
- In mosaics of XX and XO cell lines, the phenotype may vary from complete Turner syndrome to a completely normalappearing female. Likewise, in XO/XY mosaics, the phenotypic variation ranges from complete Turner syndrome to a normal-appearing (but infertile) male. How can these variations be explained?
- Suppose that part of the short arm of one chromosome 5 becomes nonreciprocally attached to the long-arm end of one chromosome 13 in the diploid set. This is considered to be a "balanced translocation" because essentially all of the genetic material is present and the phenotype is normal. One copy of the short arm of chromosome 5 produces cri du chat syndrome (mental retardation, facial abnormalities); three copies lead to early postnatal death. If such a translocation individual has children by a chromosomally normal partner, predict the (a) chromosomal and (b) phenotypic expectations.
- About 2% of patients with Down syndrome have a normal chromosome number of 46. The extra chromosome 21 has been nonreciprocally translocated onto another autosome of the D or G group. This condition is called familial Down syndrome. (a) Suppose that one phenotypically normal parent has 45 chromosomes, one of which is a translocation of the centromere and long arm of a D-group chromosome (either 14 or 15) and the long arm minus the centromere of a G-group chromosome (21). The short arms of each chromosome (presumably carrying no vital genes) are lost in previous cell divisions. If gametes from this translocated parent unite with those from a normal diploid individual, predict the chromosomal and phenotypic expectations in their progeny. (b) Assuming that in one parent the translocation is between chromosomes 21 and 22, that the centromere of the translocation is that of chromosome 22 (like centromeres go to opposite poles), and that the other parent is a normal diploid, predict the chromosomal and phenotypic expectations in their children. (c) Make the same analysis as in part (b), assuming that the centromere of the 21/22 translocation chromosome is that of chromosome 21. (d) Assuming that in one parent the translocation involves 21/21 and the other parent is a normal diploid, predict the chromosomal and phenotypic expectations in their children. (e) Among the live offspring of parts (c) and (d), what are the risks of having a child with Down syndrome?
- The photograph accompanying this problem is at the back of the book. It shows the chromosomes from a human cell. Cut out the chromosomes and construct an idiogram. Do not look at the answer until you have solutions to the following questions. (a) Is the specimen from a male or a female? (b) What possible kinds of chromosomal abnormalities may be present in this patient?
- allotetraploid (amphidiploid)
- reciprocal translocation
- position effect
- paracentric inversion
Variation in Chromosome Number
- (a) 21, triploid (b) 28, tetraploid (c) 35, pentaploid (d) 42, hexaploid
- Half of the chromosomes of hirsutum have homology with arboreum, and the other half with thurberi. Doubling the chromosome number of the sterile hybrid (thurberi × arboreum) could produce an amphidiploid with the cytological characteristics of hirsutum.
- (a) 3/8 flinty, yellow : 1/8 flinty, white : 3/8 floury, yellow : 1/8 floury, white
- (a) 11 (b) 13 (c) 14 (d) 14 (e) 10 (f) 6 ( (g)) 18 (h) 24
- (a) 2 Sss (starchy) : 1 sss (sugary) (b) 1/3 SSS : 1/6 SSs : 1/3 Sss : 1/6 sss : 5/6 starchy : 1/6 sugary
- (a) 17 yellow : 1 white (b) 5 yellow : 1 white (c) 11 yellow : 1 white (d) 3 yellow : 1 white
Variation in Arrangement of Chromosome Segments
- 26.1 map units from the point of translocation.
- First anaphase: a diad, a loop chromatid, and an acentric fragment; second anaphase: the diad splits into two monads and the loop forms a bridge. The acentric fragment formed during meiosis I would not be expected to be present at meiosis II.
Variation in the Number of Chromosome Segments
- (a) + / w and + / (–) (heterozygous deficiency) (b) 9 waltzers : 12 normals
Variation in Chromosome Morphology
- (a) All daughters have vermilion eyes (X^XY); all sons have bar eyes (XY). (b) 50% death loss; nullo-X is lethal (YY); superfemales (X^XX) usually die. (c) Same as part (a)
- (a) (1) D or C (2) B or A (b) No. Reciprocal vs. nonreciprocal exchanges are occurring in a 1 : 1 ratio, indicating that chromatids attached to the same centromere are involved in an exchange with the same frequency as chromatids attached to different centromeres. Daughters with genotype f + / + g resulting from single crossovers of type C cannot be distinguished from nonexchange chromatids. (c) Two exchanges in the garnet-forked region involving all four strands, as well as one nonreciprocal exchange between f and the centromere, are required to give homozygous wild-type and garnet-forked daughters. Their absence is support for the assumption that only two of the four chromatids undergo exchange at any one locus.
- (a) Either nondisjunction of the two X chromosomes occurred in the mother in the first meiotic division or nondisjunction of the two sister chromatids occurred in the second meiotic division. (b) Nondisjunction during the second meiotic division of the sister chromatids of the X chromosome bearing the recessive color-blind gene would produce an egg with two X chromosomes bearing only the color-blind alleles. Alternatively, if crossing over occurs between the centromere and the color-blind locus and is followed by nondisjunction of theX chromosomes at the first meiotic division, one of the four meiotic products would be expected to contain two recessive color-blind alleles.
- A sperm bearing two Y chromosomes is produced by nondisjunction of the Y sister chromatids during the second meiotic division. The other product of that same nondisjunctional second meiotic division would contain no sex chromosome; when united with a normal egg, an XO female with Turner's would be expected.
- If mitotic nondisjunction occurs early in embryogenesis, mosaicism is likely to be widespread throughout the body. If it occurs late in embryogenesis, mosaicism may be limited to only one organ or to one patch of tissue. If chromosomally abnormal cells are extensive in reproductive tissue or in endocrine tissues responsible for gamete and/or sex hormone production, the effects on sterility are likely to be more intensively expressed.
- (a) 1 normal karyotype : 1 balanced translocation : 1 deficient for short arm of chromosome 5 : 1 with three copies of the short arm of chromosome 5 (b) 2 normal : 1 cri du chat syndrome : 1 early childhood death
- (a) 1 chromosomally and phenotypically normal (2n = 46) : 1 translocation carrier, phenotypically normal (2n – 1 = 45) : 1 monosomic (2n = 45) for a G-group chromosome aborted early in pregnancy) : 1 translocation Down trisomic for the long arm of chromosome 21 (2n = 46). Among the live-born offspring we expect 1/3 chromosomally normal : 1/3 translocation carriers : 1/3Down syndrome. (b) 1 chromosomally and phenotypically normal (2n = 46) : 1 that is a 21/22 translocation carrier, phenotypically normal (2n – 1 = 45) : 1 monosomic for chromosome 21 and aborted early in pregnancy (2n – 1 = 45) : 1 with a 21/22 translocation chromosome who is essentially trisomic for the long arms of 21 (2n = 46) and phenotypically Down syndrome. (c) 1 chromosomally and phenotypically normal (2n = 46) : 1 that is a 21/22 translocation carrier, phenotypically normal (2n – 1 = 45) : 1 monosomic for 22 and aborted early in pregnancy (2n – 1 = 45) : 1 with a 21/22 translocation chromosome who is essentially trisomic for the long arms of 22 (2n = 46), phenotype unspecified. (d) 1 monosomic (2n – 1 = 45) for 21 and aborted early in pregnancy : 1 with a 21/21 translocation chromosome who is essentially trisomic for the long arms of 21 and phenotypically Down. (e) 1 in 3 for part (c); 100% for part (d).
- (a) Male (b) The karyotype contains an extra G-group chromosome (2n + 1 = 47). It cannot be determined whether the extra chromosome is 21, 22, or Y. If the patient has the physical characteristics of Down syndrome, the extra chromosome is 21.
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