BIO 325 - Fall 2009 - TTh 930 to 1100 - Exam 1 - Key

BIO 325 - Fall 2009 - TTh 930 to 1100 - Exam 1 - Key -...

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GENETICS (BIO 325) Fall 2009 T, Th 9:30 – 11:00 AM section Key - Exam 1 __________________________________________________________________________________ 1. (a) Autosomal recessive (b) 1/16 2. Both parents must be heterozygotes ( R/r ). Offspring with black eyes( R /-); offspring with red eyes ( r/r ) 3. (a) T/T × t/t T/t (F 1 ); T/t × T/t F 2 progeny = 1( T/T ), 2 ( T/t ), 1 ( t/t ) = 3:1 (tall:short). For the tall plants, two genotypes exist ( T/T and T/t ), therefore selfing the F 2 tall plants will yield two groups of progeny; T/T × T/T crosses will produce only tall progeny, whereas T/t × T/t crosses will generate the same 3:1 ratio (tall:short) observed in the F 2 progeny. Overall, then, you expect a 7:1 tall:short ratio. (b) Selfing the short F 2 progeny ( t/t ) will only yield short progeny. 4. (a) Because the first child is albino, both parents must be carriers for the recessive albino allele. The probability of each child having the albino genotype a/a equals 1 4 = 0.25 = 25%. Using the product rule, the probability of the next two children being albino is ( 1 4 ) ( 1 4 ) = 1 16 = 0.0625 (6.25%). (b) The probability of the couple having any child with normal pigmentation (genotypes A/ A or A/a ) is 3 4 (75%).
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This note was uploaded on 05/15/2010 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas.

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BIO 325 - Fall 2009 - TTh 930 to 1100 - Exam 1 - Key -...

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