Homework 2 Key

# Homework 2 Key - HOMEWORK SET#2 KEY 1(6 pts total a The...

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HOMEWORK SET #2 — KEY 1. (6 pts total) a) The minimum number of genes is 3 to specify 8 different phenotypes. If one assumes that a different phenotype is due to A_ vs aa; B_ vs bb, etc., there will be 2 3 combinations (A_B_C_ vs aaB_C_, A_bbC_, etc. (2 pts) b) This is one example that will fit the data, perhaps others are possible. (4 pts) Youngberry aaBbCc Santiam Aabbcc Loganberry AaBbCc NW Native Blackberry AabbCc Trailing Blackberry AabbCc Wild Himalaya AaB_cc Marionberry aabbcc Chehalem aaBbcc Olallieberry aabbCc 2. ( 10pts total ) a) number of monozygotic twins = 7,000 number of dizygotic twins = 3,000 10,000 twin births 4,350 ¢ ¢ pairs 4,150 pairs 1,500 ¢ pairs To find the number of monozygotic and dizygotic twins in the population, you must first look at the ¢ pairs as they are all the product of two eggs, each fertilized by a separate sperm. ( 4 pts.) Look at all possible dizygotic twin pairs ¢ ¡ ¢ ¢ ¡ ¢ ¡ ¢ ¡ : ¢ – ™ ¢ – ¢¢ ¡ ¢ ¢ 1 2 1 : : : • male-female pairs = 1/2 of all possible combinations • you may then conclude that 1,500 ¢ twins represents 1/2 of total dizygotic twin pairs 3,000 total number of dizygotic twins • 750 : 1,500 ¢ : 750 ¢ ¢ o ¢ ¢ ¢ Total

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Total # 4,350 1,500 4,150 10,000 Dizygotic 750 1,500 750 3,000 Monozygotic 3,600 3,400 7,000 b) ( 4 pts. ) p (dizygotic twins having identical phenotype) = 15 128 Autosomal genes ¢ Aa Aa A>a MN MN M = N Bb bb B>b
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