See pedigree info on extra sheet 1. ( 6 pts)

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1 Biol 321 Quiz #2 Spring pts NAME See pedigree info on extra sheet 1. ( 6 pts) Examine the pedigree shown above. For each mode of inheritance listed below indicate: E = this mode of inheritance is excluded by the data C = this mode of inheritance is consistent with the data C Autosomal recessive inheritance; recessive allele is common in the population ARC E Autosomal recessive (recessive allele is very rare) ARR (all spouses need to be hets) E X-linked recessive (recessive allele is common) XRC (affected female has unaffected dad) E X-linked recessive (recessive allele is very rare) XRR (affected female has unaffected dad) E X-linked dominant (dominant allele is very common) XD (affected male has unaffected mom frequency of allele is not relevant in this case) C Autosomal dominant AD For each mode of inheritance that you excluded, circle the portion of the pedigree that excluded this inheritance pattern and label with the appropriate acronym (indicated by letters in bold). Note, you do not need to explain why the region of the pedigree excludes the mode of inheritance-- just circle and label it. If more than one region of the pedigree excludes a particular mode of inheritance, just indicate one region. Be as precise as possible when indicating the portion of the pedigree that is relevant. 2. (2 pts.) Superman is a gene found in Arabidopsis thaliana (and many other plant species). It plays a role in controlling the boundary between stamen (male reproductive parts) and carpel (female reproductive parts) development in a flower. Briefly speculate (1-2 sentences max) as to why this gene is called superman and not superwoman. Genes are often named for the mutant phenotype that first brought the existence of the gene to the attention of a researcher. In this case, loss-of-function mutations result in an excess of stamens hence the name superman. By the way, there is an Arabidopsis gene named ClarkKENT 1

2 3. ( 5 pts.) Two individuals of genotype Aa mate and produce 5 kids. a. (2 pt.) What is the probability that all 5 kids are heterozygous? (1/2) 5 b. (3 pts.) As it turns out, all 5 kids show the dominant phenotype. Given this information, what is the probability that (i) all 5 kids are heterozygous (2/3) 5 (ii) some of the 5 kids are het (Aa) and others are homozygous (AA) [Just set up answer & make sure your logic is clear] 1- [(1/3) 5 + (2/3) 5 ] = all possible outcomes with the exception of all homozygous or all heterozygous 4. (2 pts) The current model for the evolution of the Y chromosome suggests that the mammalian X and Y chromosomes were a regular pair of autosomes 300 mya (million years ago). What evidence was presented in lecture in support of this assertion? 1-2 sentences max The X and Y chromosome share genes at both chromosomal tips. Furthermore, sprinkled along the length of the Y chromosome are a handful of genes that are also found on the X representing remnants of the Y chromosomes former stature as a more regular chromosome. 5. (1 pt.) Recall the article entitled: Zebrafish researchers hook gene for human skin color. Based on this article, indicate whether each statement is true or false. Answer false if any part of the statement is false. FALSE Allelic variation in this single gene accounts for over 99% of the skin color variation between individuals of European and African descent. 6. ( 2pts) By each statement circle True/False/Not addressed in the Komodo dragon paper. Answer false if any part of the statement is false. TRUE Reproduction in Komodo dragons is plastic in the sense that females can switch back and forth between parthenogenesis and normal sexual reproduction. TRUE The authors speculate that parthenogenesis is adaptive because a single unfertilized female could found a colony by mating with her parthenogenetic offspring. 2

3 7. (6 pts.) Recall Problem 5 from Quiz 1 Each line represents a single polymer of double-stranded DNA. (The DNA polymers will segregate as indicated.) Fill in the blanks in parts a and b: a. (1 pt) This diagram is consistent with a 1n = 2 cell undergoing mitosis b. (1 pt) This diagram is also consistent with Meiosis II in a 2n = 4 organism of genotype AaBb c. (1 pt.) For part b above, what fraction of the all of the Meiosis II divisions occuring in the gonad of the organism would resemble the diagram? [Ignore complications from crossing over events.] ¼ d. (3 pts) Illustrate your answer to part c by showing all other meiosis II divisions in an organism of genotype AaBb. [Ignore crossing over events.] USE the same schematic form of the above diagram. Be sure to label alleles but don t include any other details. Three diagrams: showing the Ab, AB and ab combos Problem (pts) 1-2 (8 ) 3-6 (10) 7 (6) 8 (8) 9 (8) Total (40 pts.) Score 3

4 8. ( 8 pts.) Cranio frontonasal syndrome (CFNS) is a very rare birth defect in which premature fusion of the cranial structures leads to abnormal head shape, widely spaced eyes, nasal clefts and various other skeletal abnormalities. George Feldman and his colleagues looked at several families in which CFNS occurred and recorded the results shown in the following table (G.J. Feldman, 1997, Human Molecular Genetics 6: ) Examine the table carefully. Consult the pedigree rules on the extra sheet. a. (1 pt) Is the CFNS mutation dominant or recessive to the wildtype allele? No explanation necessary. dominant b. (3 pts.) Let s examine Family 7b as an isolated set of data. Assign appropriate allele symbols and and write out genotypes of the parents assuming the CFNS gene is located on an autosome. THEN calculate the probabililty of these parents producing two affected daughters. Allele symbols: C = CFNS allele c = normal (wt) allele Father s genotype Cc Mother s genotype cc Probability these parents would produce two affected daughers: for each: ½ female X ½ Cc Overall probability of 2 affected daughters = 1/16 b. (3 pts.) Again, let s examine Family 7b as an isolated set of data. Assign appropriate allele symbols and and write out genotypes of the parents assuming the CFNS gene is located on the X chromosome. THEN calculate the probabililty of these parents producing two affected daughters. Allele symbols: C = CFNS allele c + = normal (wt) allele Father s genotype X C Y Mother s genotype X c+ X c+ Probability these parents would produce two affected daughers: for each = ½ (probability that child received the X chromosome from dad) Overall probability of 2 affected daughters = 1/4 c. (2 pts) Taking all of the data together, what do they suggest about the location of the CFNS gene in the human genome? Choose the best answer c. likely to be in the differential region of the X chromosome 4

5 9. (8 pts.) Consult page 2 of extra sheet for Sex Determination Table, background information on the Na vi and details of the traits under examination. For this problem assume parents are from true-breeding subpopulations. Crosses are repeated on the extra sheet for your genotyping convenience a. A female Na vi with a narrow nose bridge mates with a male Na vi with a wide nose. All of their daughers have wide noses and their sons have narrow noses. For each mechanism of sex determination listed below, indicate whether it is Consistent with or Excluded by this cross. If excluded, explain why with genotypes or in one complete sentence. XX/XY Circle: Consistent ZZ/ZW Circle: Excluded If excluded, explain why Would have expected all sons and daughters to have wide noses since DAD would be homozygous for the dominant allele mom would be hemizygous for the recessive allele Haploid/Diploid Circle: daughters are heterozygous Consistent -- sons are haploid and have mom s recessive allele; b. The same parents also differ with respect to digit length. The female has long digits and her mate has short digits. All of their offspring have digits of medium length. Does this information influence your assessment of the sex determination mechanism that operates in the Na vi? Circle: YES, haplodiploidy is eliminated by this observation Briefly explain your answer in one or two sentences. All offspring show an intermediate phenotype and therefore are heterozygous. Since males clearly have two gene copies, they cannot be haploid. 5