Prelim+1+F23+KEY

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BioMG2800 © Department of Molecular Biology and Genetics Cornell University 1 BIOMG 2800 Lectures in Genetics and Genomics First Preliminary Exam September 12, 2023 There are 11 pages to this exam, including this page. Be sure you have a complete set before you begin. Pages 10 and 11 are blank pages at the end. You may remove these and use them for scratch paper (you do not need to turn them in). Put all answers in the boxes or on the lines provided. Only answers in the boxes or lines will be graded. Please turn off all cell phones and other electronic devices. You may use a non-internet capable calculator. ___Answer Key_______ _______________ Name: LAST, FIRST Net ID 1. ________/10 2. ________/12 3. ________/20 4. ________/20 5. ________/20 6. ________/18 Total ________/100

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 2 1. (10 Points) For the following statements, indicate whether they are true or false. XY chromosomes are not homologous to each other, but they can still pair during meiosis I, and undergo cross-overs (via the PAR regions). T For any given gene, the only case when parents (maternal and paternal) do not contribute equally to their offspring is when that gene is linked to another gene on the same chromosome. F When a single nondisjunction event occurs in meiosis, all 4 resulting gametes will always have an abnormal number of chromosomes F In humans, X-chromosome inactivation induces dosage compensation by randomly inactivating one of the X chromosomes in females. T Genes B and R are 20 map units apart. In an individual with genotype B r / b R , a gamete with genotype B R will be produced 10% of the time T

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 3 2. (12 points) A farmer conducted several rounds of breeding to expand their Highland cow stocks. The cows had various hair lengths: long, medium, short, or curly. Each breeding scheme started with several pairs of true-breeding Highland cows of the hair types indicated in the table below. Each set of pairings yielded identical F1s of a single hair type (not recorded). The F1s from each breeding pair were then intercrossed (F1 X F1) to produce the F2 offspring with the hair types indicated in the table below. Original breeding pairs F2 Offspring (number) Long x Medium Long (15), Medium (42) Medium x Short Medium (33), Short (12) Medium x Curly Curly (15), Medium (45) Long x Short Long (27), Short (9) Short x Curly Curly (12), Short (39) Long x Curly Long (45), Curly (12) Using the data above provide a genetic explanation for how hair type is determined. For your answer, indicate (i) how many gene(s), (ii) how many alleles for each gene, and (iii) the dominant/recessive relationship(s) between all allele(s) for each gene involved. Use notation like A 1 > A 2 or A > a to indicate which allele is dominant. If two alleles are co/incompletely dominant to each other, use notation like A 1 = A 2 , etc. 1 gene (A) + 1 pt with 4 alleles (A l , A m , A s , A c ) + 2 pts A L = long, A M = medium, A S = short, A C = curly A M > A L > A S > A C +9 pts for the allelic relationships:

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BioMG2800 © Department of Molecular Biology and Genetics Cornell University 4 3. (20 points) You have identified a new species of mongoose and determined that they are a diploid organism. You also determine this species uses the same mechanism of sex determination as humans (where females are homogametic and males are heterogametic). The five cells shown below (A-E) represent different cell types from the same animal. Assume, for this question, that nondisjunction (NDJ) only occurs during meiosis and that only a single NDJ event would occur. i. (3 pts) What is N for this organism? ____ 3 _____ ii. (2 pts) Is this animal a male, a female, or cannot determine? ___ female _____ iii. (15 pts) Match the single best description to each of the cells. 1. A cell in G1 2. A cell after DNA replication (before meiosis or mitosis) 3. A cell in metaphase of mitosis 4. A cell in metaphase of Meiosis I 5. A cell in metaphase of Meiosis II 6. A normal gamete 7. A gamete resulting from a single non-disjunction in meiosis I 8. A gamete resulting from a single non-disjunction in meiosis II. 9. A gamete resulting from a single non-disjunction in meiosis I or II (can’t tell). Cell A 2 Cell B 6 Cell C 9 Cell D 9 Cell E 1

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 5 4. (20 points total) Pedigrees. Pedigree 1: Consider the above human pedigree for a rare disorder causing dry eyes. Assume complete penetrance. a. (4 pts) What is the most likely mode of inheritance for this disorder? Autosomal dominant, autosomal recessive, X- linked dominant, or X-linked recessive. Autosomal recessive b. (4 pts) What is the probability that individual III-6 is a carrier? 1/2 c. (5 pts) Suppose individuals IV-2 and IV-3 had a child. What is the probability that this child will have the disease? If the disease probability differs between a male child and a female child, list the probabilities separately. 1/16 same for male / female Prob IV-2 is a carrier = 1 Prob IV-3 is a carrier = ¼ (½ that III-6 is a carrier * the ½ chance III-6 will pass the disease allele to IV-3) ¼ chance the child will be homozygous recessive when both parents are carriers.

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 6 4. Pedigrees continued Pedigree 2: Consider the above human pedigree for a rare disorder causing flaky skin. Assume complete penetrance. d. (4 pts) What is the most likely mode of inheritance for this disorder? Autosomal dominant, autosomal recessive, X- linked dominant, or X-linked recessive. X-linked dominant +4 Autosomal dominant +1 e. (3 pts) Indicate if individuals III-3, III-4, and IV-4, are affected, unaffected, or unable to be determined. III-3 ___ unable to determine _________ III-4 ___ unaffected ________ IV-4 ___ affected _________ If autosomal dominant written in part d, answers will be: III-3 = unable to determine III-4 = unable to determine IV-4 = unable to determine

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BioMG2800 © Department of Molecular Biology and Genetics Cornell University 7 5. (20 points) The Micronesian Kingfisher, Halcyon cinnamomina , has a cinnamon-colored face. In some kingfishers, the color continues onto the chest, producing one of three patterns: a circle, a shield, or a triangle. In other kingfishers, there is no cinnamon color on the chest. A male with a triangle was crossed with a female with no color on her chest. All their F1 offspring had a shield on the chest. The F1 offspring were crossed with each other, and they produced the following F2 birds: 66 circle: 128 shield: 62 triangle : 84 no color. a. (2 pts) How many genes are controlling this trait:___ _2 ____ b. (18 pts) Indicate the genotypes of the birds in all three generations of this cross: +2 Male triangle:____ A 1 A 1 BB ___ X +2 Female no color: _____ A 2 A 2 bb ____ Genotypes must be internally consistent. Triangle and no color genotypes below must agree with what is written for parents. Progeny must be consistent with each other. (Any symbols consistent with how nomenclature was discussed in class can be used). +2 F1 shield:___ A 1 A 2 Bb ___ F2: +3 Circle:__ A 2 A 2 B- __ +3 Shield:__ A 1 A 2 B- __ +3 Triangle:__ A 1 A 1 B- __ +3 No color:__ -- bb __

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 8 6 . (18 points total) True-breeding male hamsters with bubbly ears were mated to true-breeding females with glassy eyes, harpoon whiskers, and extra toes. The F1 generation consisted of males and females with harpoon whiskers (eyes, ears, and toes were wildtype). The harpoon whiskered F1 females were crossed to males with bubbly ears, glassy eyes, and extra toes. 1000 F2 progeny were scored as indicated in the table below. Only the mutant phenotype (and not the body part) is written. Number F2 Phenotypes (males and females) 155 glassy, harpoon 162 glassy, extra, harpoon 164 bubbly 157 bubbly, extra 60 glassy, bubbly, harpoon 57 glassy, bubbly, extra, harpoon 61 wildtype 56 extra 30 glassy, extra 32 glassy 27 bubbly, harpoon 32 bubbly, harpoon, extra 1 glassy, bubbly, extra 2 glassy, bubbly 2 harpoon 2 extra, harpoon 1000 Total F2 progeny

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 9 a. (4 pts) Fill out the table below for the genes involved in this cross. Indicate whether the mutant allele is dominant or recessive to the wildtype allele. Indicate whether the gene is X- linked, Y-linked, or autosomal. Gene Dominant or recessive X-linked, Y-linked, or autosomal? bubbly recessive autosomal glassy recessive autosomal extra recessive autosomal harpoon dominant autosomal b. (14 pts) Diagram the chromosomes of the F1 females, using horizontal lines to indicate the chromosomes. Show each chromosome as a single line. • Indicate all alleles of all genes in your diagram. Use G and g for the glassy gene alleles, B and b for the bubbly gene alleles, E and e for the extra gene alleles, and H and h for the harpoon gene alleles. • If any of the genes are linked, give map distances (expressed to one decimal place).

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Prelim+1+F23+KEY

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