GENETICS 310 PRACTICE EXAM I-1 ANSWERED I The results of four crosses are shown below. Put a legend for the inheritance of each character in each problem in the "Legends" box, and then use your legend to assign genotypes to the parents and progeny in the appropriate blanks. Cross 1. (P 1 ) Green X (P2) yellow (both are truebreeding) P 1 GG P 2 gg G_ -green gg - yellow Legends F 1 ; all green Gg F 2 ; 3/4 green : 1/4 yellow 2. (P 1 ) Green X (P 2 ) Green (both truebreeding) P 1 Y1Y1,y2y2 F 1 ; all yellow Y1y1, Y2y2 F 2 ; 9/16 yellow : 7/16 green 4 P 2 y1y1, Y2Y2 3. (P 1 ) Black X (P 2 ) white (both true breeding) P 1 B"B" F 1 : All Blue F2: 1/4 Black B"B" P 2 BB B"B 4. Stumpy (P1) by Stumpy (P2) F1: 6 Stumpy S'S: 3 Long SS :2/4 Blue B"B :1/4 white BB Predict the progeny of this cross: Stumpy F1 by Long F1 Y1_ - yellow Y2_ Yellow y1y1 - green y2y2 green with epistasis: or: Y1_, Y2_ Yellow y1y1, Y2_ -green Y1_, y2y2 -green y1y1, y2y2 -green B"B" - black B"B -blue BB - white S'S' - lethal S'S - stumpy SS - long Stumpy 1 : Long 1 1
II Myotonic dystrophy in humans is inherited as a dominant trait. Infants who are affected generally have cataracts, very small gonads and extremely weak muscles. In an adult-onset form, symptoms usually begin with muscle spasms (myotonia) leading to weakness, frontal balding and cataracts often develop. Family pedigrees typically show either the infantile or adult form. In some cases, progeny of a mildly affected parent may show no symptoms but have a child with severe symptoms. Since many of the symptoms also occur in other genetic diseases or as a result of non-genetic causes, ECG tests can be used to confirm the diagnosis. Which of the following terms can be applied to myotonic dystrophy? Tell what aspect of the description led to your "positive" decisions: Term Aspect X pleiotrophy X age of onset X genetic heterogeneity multiple organs/phenotypes infant and adult forms other genetic diseases cause same symptoms; two forms multiple alleles X_ lack of penetrance; X_variable expressivity: cases unaffected progeny of affected parent pass it on pass the trait on different degrees of expression, mild and severe lethal gene _X phenocopy: non genetic causes produce similar phenotypes teratogen epistasis _ quantitative trait 2
III. In cats: M"M" -Lethal (unborn) Pk_ -polycystic kidney S_ -Short hair P_ -6 toes M" M -Manx (tailless) pk/pk -normal kidney ss -long hair pp -5 toes M M -normal A Manx female with polycystic kidney, short hair and 5 toes is mated to a male that is also tailless (ie Manx), has polycistic kidney, short hair but has 6 toes. The first litter has kittens with each homozygous recessive phenotype. A Give the genotype of the female: M'M, Pk/pk, Ss, pp B) give the genotype of the male: M'M, Pk/pk, Ss, Pp C) How many genetically different kinds of gametes can the male produce? 2 X 2 X 2 X 2 = 16 D) If the same parents have another litter, what is the probability of a M"M, Pk/pk, SS, Pp kitten? 2/3 X 1/2 X 1/4 X 1/2 E) What fraction of the kittens are expected to be Manx, with polycystic kidney, long hair and 5 toes? 2/3 X 3/4 X 1/4 X 1/2 F) Two Manx cats are crossed. (1) If 6 kittens are born, what is the probability 4 will be Manx and 2 normal? (6!/4!2!) (2/3) 4 (1/3) 2 (2) What is the probability that at least one of the 6 kittens will be Manx? 1-(1/3) 6 (the only case is where all 6 are normal) 3
IV. Females from a short-eared (8 cm) rabbit breed were crossed to a male from a long-eared breed (16 cm). A) Assuming all parents are true breeding and that ear-length is a multi-gene trait, what ear-length is expected in the F1? 12 B. F2 Progeny from sib matings of F1 males and females have ears that range from 8 to 16 cm in length; 1) Does transgressive segregation occur in this problem? No Explain your answer F2 extremes do not surpass parents in P1 2) F2 rabbits were found with 8, 9, 10, 11, 12, 13, 14, 15 and 16 cm long ears. How many "ear-length" genes are heterozygous in the F1? 2N+1 = 9 classes, so N = 4 3) What fraction of the F2 progeny will have 8 cm ears? (1/4) 4 4) Propose genotypes for each of the original P1 rabbits. AA, BB, CC, DD doe by A'A', B'B', C'C', D'D' buck C. If the Variance in the F1 generation is 2 and in the F2 generation it is 10, what is the value for H 2? (Show your derivation, including the various sources of variation). V T = 10 from F 2 (gene and environment affect ear length) V E = 2 from F1 (all are A'A, B'B, C'C, D'D, so only Environmental effects) Vg = Vt - Ve = 8 H 2 = Vg/Vt = 8/10 = 0.8 VI. Explain the rationale for using twins to investigate the heritability of human traits. ID = MZ twins have the exact same genotypes, so any differences show Ve; DZ twins can differ by genes as well as environment, so estimate Vt. Twins reared apart can maximize Ve differences. VII An error at the hospital left 4 baby boys born the same day with unlabeled wristbands. Since mother and baby blood types were accessible, some sorting could be done. Although the fathers were not available, 4 sufficient information was gained to assign babies toa mother
A, M, Rh + Mom -A Mom-B Mom-C Mom-D Baby O, N, Rh + A, M, rh - AB, N, Rh + O, MN, rh - 1 MN MN 2 O, N, rh - MN ABO 3 AB, M, Rh + ABO, MN MN ABO 4 B, MN, rh - B) Which baby will you send home with each mother? A 2 B 3 C 4 D1 C) Which mother(s), if any, should have a shot of Rhogam? Why B & D are both rh - but have Rh + babies so will make anti Rh unless treated VII. An unusual organism has only 2 chromosomes, one long and one short in its gametes. Show how the chromosome would align in metaphase of: Mitosis Meiosis I 5