Exam 1 (1)

Sp/Su 2011 Name:_ Exam 1 May 26, 2011 Access ID: Tools you may need: Section 1. Multiple Choice (2.5 points each, 40 points total). Clearly circle the letter of the correct answer. 1) Mendel crossed a two yellow, short plants and received the following offspring: 0 yellow, tall 42 yellow, short 0 green, tall 25 green, short What are the genotypes of the parents assuming Y = yellow allele, y = green allele, T = tall allele, t = short allele? a) yytt b) Yytt c) YYTT d) YyTT e) yyTT 2) How many sperm cells will form from fifty primary spermatocytes? Primary spermatocytes are 4n, therefore expect 4*50 = 200 3) The Purple People Eater has 5 distinct chromosomes (2n = 10) and 40 known genes, each with two alleles. If an individual is heterozygous at 50% of all known loci, and 50% homozygous recessive at the remainder of the loci, approximately how many different gametes can be produced if all genes independently assort? a. 1x106 b. 1x1012 c. 1x1015 20 of the genes do not matter since homozygous, 20 6 d. 1x101 e. 1x103 other 20 assort independently therefore 2 = 1.05x10 4) Assuming a 1:1 sex ratio, what is the probability that 4 children produced by the same parents will be born with the sexes alternating?  (1/2)4  + (1/2)4  = 0.125 Two possible outcomes, order matters – sum law states: two independent events occurring but the combination can occur in more then one way, therefore add the probability of each event happening _Points for Page v1 2 Sp/Su 2011 Name:_ Exam 1 May 26, 2011 Access ID: 5) How many chromatids are present at each stage of the cell cycle in a cell with eight pairs of chromosomes? Stage of Cell Cycle: G1 G2 G0 G2/M 6) How many degrees of freedom will there be for a c2 test of the phenotypes that appear in the F2 generation following a P1 cross of four genes between pea plants that differ for Mendel’s genes assume the following: 1) the initial parents were true breeders, 2) dominance is complete and 3) the traits are not linked? 16 Possible phenotypes (24) Df = n-1 = 16-1 = 15 7) In the cross above (question #6), what proportion of the offspring are expected to be homozygous dominant for two of the genes and homozygous recessive for the other two? (1/4)(1/4)(1/4)(1/4) = 1/256 8) Which of the following diagrams show homologous pairs aligned during mitosis (black = maternal; white = paternal)? 9) Which of the following statements about Prophase I of Meiosis is incorrect? a. During leptonema, homologous chromosomes align and pair. (Pairing is not yet complete!) b. After diplonema, crossing over is complete. c. During zygonema, synapsis is initiated. d. During pachynema, meiotic recombination sites mature. e. During the process of synapsis, homologous pairs are held together by cohesion and crossing over. 10) Model organisms are essential to the study of genetic and physiologic mechanisms. Which of the following is not a characteristic of a model organism? a. Relatively short life cycles b. Easy to grow and maintain. c. Mechanisms are always the same across species (Mechanisms are the same across most, but not all species) d. Produce many offspring e. Discrete variations are available _Points for Page v1 3 Sp/Su 2011 Name:_ Exam 1 May 26, 2011 Access ID: 11) In what stage of meiosis does an error most likely to account for presence of Down’s syndrome (three copies of chromosome 21) in humans, where homologous pairs fail to separate? a. Anaphase I b. Anaphase II c. Prophase I d. Prophase II e. Metaphase I f. Metaphase II 12) Which of the following modes of inheritance describes an interaction that occurs between multiple alleles to affect the phenotype? a. Epigenesis b. Pleiotropy c. Epistasis d. X-linked e. Codominance (all other options are interactions between genes, except X-linked which is sex specific) 13) In the ABO blood system in human beings, alleles IA and IB are codominant, and both are dominant to the IO allele. In a paternity dispute, a type AB woman claimed that one of four men, each with a different blood type, was the father of her child with type B blood. Which of the following could be the blood type of the father of the child on the basis of the evidence given? a. Type O b. Type AB c. Type A d. Type B e. All of the above f. None of the above 14) In the pedigree below, what is (are) the possible mode(s) of inheritance? Select the best answer. a. X-linked dominant b. Autosomal dominant c. X-linked recessive d. Autosomal recessive e. A or B f. C or D  X-linked possible because occurs in crisscross pattern, mostly in males Autosomal recessive possible because it skips generations 15) A researcher isolates two types of mutants from pea plants that do not have flowers. Which of the following analysis should the researcher use to determine if the mutations occurred in the same gene? a. Dihybrid Cross Analysis b. Testcross Analysis c. Pedigree Analysis d. c2 Analysis e. Complementation analysis If cross two mutants together, and offspring are mutant then may be two different mutations in same gene. If offspring are normal, then mutations in two separate genes that affect same phenotype. _Points for Page v1 4 Sp/Su 2011 Name:_ Exam 1 May 26, 2011 Access ID: 16) A scientist was measuring the percentage of nitrogenous bases in 5 samples. Which of the following molecules is DNA? Nucleic Acid Molecule %A %T %G %C %U a. 28 28 22 22 0 b. 31 0 31 17 21 c. 15 35 15 35 0 d. 26 0 24 24 26 e. 30 15 20 20 15 Section 2. Short answer. (Points indicated for each question). Write clearly in the space provided. Define any symbols you use in your calculations and show all work. 17) List and briefly describe Mendel’s Postulates of Inheritance. Provide supporting data from his initial experiments that led him to these conclusions (12 points). · Unit Factors in Pairs (1 point): Genetic characters are controlled by unit factors existing in pairs (1 point). Each trait studied only had two possible outcomes (alleles) (1 point). · Dominance/Recessiveness (1 point): When two unlike unit factors are responsible for a single character are present in a single individual, one unit factor is dominant to the other, which is said to be recessive (1 point). In each monohybrid cross, the trait expressed in the F1 generation was controlled by the dominant unit factor. The trait that was not expressed was recessive. Recessive traits reappeared in F2 generation (1 point). · Segregation (1 point): During the formation of gametes, the paired unit factors separate or segregate, randomly so that each gamete receives one of the other with equal likelihood (1 point). If an individual contains a pair of like unit factors, then all its gametes receive one of that same kind of unit factor. If an individual contains unlike unit factors, then each gamete has a 50:50 chance of receiving either kind of unit factor (1 point). · Independent Assortment (1 point): During gamete formation, segregating pairs of unit factors assort independently of each other (1 point). Transmission of one pair of unit factors did not influence the outcome of segregation of any other pair (1 point). 18) How do we know that DNA is the Genetic Material? Outline and describe the key experiments that led to the acceptance of DNA as the genetic material (6 points). Answer should mention (1 point per names, 1 point per outcome of experiment): • Griffith – heat killed virulent bacteria were able to “transform” live avirulent bacteria into virulent bacteria • Avery, MacLeod, and McCarty – Showed that DNA was the factor responsible for “tranformation” through systematic elimination of other chemical components of the cell (Bonus points – 2 if they describe all the steps of the experiment) • Hershey-Chase Experiment – DNA was the component that viruses injected into bacteria to make more virus (Bonus Points – 2 if describe radiolabelling of protein with 35S and DNA with 32P) _Points for Page v1 5 Sp/Su 2011 Name:_ Exam 1 May 26, 2011 Access ID: 19) Clearly label on the figure the following aspects of DNA (10 points): a. Major Groove b. Minor Groove c. and d. Two features identified by X-ray diffraction e. Hydrophobic region f. Hydrophilic region g. Region of Hydrogen bonding h. Direction of strands i. Hydroxyl group j. Phosphate group 20) A troll has a chromosome number (n) of 12 while a goblin has a chromosome number of 20. Sterile hybrid troll-goblins can be produced (2 points). a. What would be the expected chromosome number in the somatic cells of the hybrids? 20 + 12 = 32 (1 point) b. Assume that the G1 nuclear DNA content of goblin is 30 pg and that the G1 nuclear content of the troll is 18 pg. What would be the expected DNA content in a metaphase somatic cell of the hybrid? 48 pg (1 point) – goblin gamete would have 15 pg, troll gamete would have 9 pg, a zygote would have 24 pg, therefore after S phase would have 48 _Points for Page v1 6 Sp/Su 2011 Name:_ Exam 1 May 26, 2011 Access ID: 21) Inheritance of a Widow’s peak is a dominant trait (W). The recessive phenotype to this trait is a straight hairline (w). For the following families, determine the genotypes of the parents and the expected phenotypic ratio of offspring. Also, determine the genotypes of the offspring and the probability of the specified combination of children occurring. If multiple genotypes are possible, list all possible combinations (16 points). a. Two parents with Widow’s peaks have five children, four with straight hairlines and one with a Widow’s peak. Parent genotypes: Ww x Ww (2 points – one for each parent) Expected phenotypic ratio of offspring: 3 straight hairline: 1 Widow’s peak (1 point) Genotypes of offspring: Four with straight hairlines ww, one with Widow peak could be WW or Ww (2 points – one for each phenotype) Probability of combination of children occurring: MUST DEFINE TERMS (1 point) If a = probability of straight hairline, b = probability of Widow’s peak: 5a4b (1 point) = 5(1/4)4(3/4) = 0.015 (1 point) b. One parent with a straight hairline and one parent with a Widow’s peak have three children, two with straight hairlines and one with a Widow’s peak. Parent genotypes: ww x Ww (2 points – one for each parent) Expected phenotypic ratio of offspring: 2 Widow’s peak: 2 straight hairline (1 point) Genotypes of offspring: two with straight hairlines ww, one with Widow’s peak Ww (2 points) Probability of combination of children occurring: MUST DEFINE TERMS (1 point) If a = probability of Normal hairline, b = probability of Widow’s peak: p = 3a2b (1 point)= 3(.5)2(.5) = 0.375 (1 point) _Points for Page v1 7 Sp/Su 2011 Name:_ Exam 1 May 26, 2011 Access ID: 22) In rabbits, a series of multiple alleles controls coat color in the following way: C is dominant to all other alleles and causes full color. The chinchilla phenotype is due to the cch allele, which is dominant to all alleles other than C. The ch allele, dominant only to ca(albino), results in the Himalayan coat color. Thus, the order of dominance is C > cch >ch>ca (14 points). a. In a P1 cross of an albino with a chinchilla results in at least one albino offspring. In a second P1 cross, a full color offspring is obtained from full color and albino parents. Determine the genotypes of the P1 generation and the F1 offspring. First Cross: Albino genotype caca (1 point) Chinchilla phenotype cchca (1 point) Albino offspring genotype caca (1 point) Second Cross: Full color genotype C_ (1 point) – impossible to determine genotype Albino genotype caca (1 point) Full color offspring Cca (1 point) b. If the F1 albino from the first mating is crossed with the F1 full color rabbit from the second mating, what would be the predicted phenotypic ratio of the F2 offspring. State as a null hypothesis. Offspring from a cross of caca x Cca would result in ½ full color and ½ albino. (1 points) c. In the F2 generation from this cross, the rabbits had 3 litters of 8 bunnies resulting in 19 full color and 5 albino offspring. Using the c2 test, determine if these results are expected (assume significant if p ? 0.05). Need to show set up with observed and expected numbers (2 points) Total bunnies = 24 Observed Expected Deviation Full color 19 .5(24) = 12 19-12 = 7 Albino 5 .5(24) = 12 5-12 = -7 c2 = (deviation2/expected)full color + (deviation2/expected)albino = (72/12) + (-72/12) = 4.08 + 4.08 = 8.17 (3 points, 2 for set up of equation, one for correct value) Df = n-1 = 2-1 = 1 (n = 2) (1 point) According to chart, p is between 0.01 and 0.001. Therefore, reject the null hypothesis in b. (1 point) _Points for Page v1 8