{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

This preview shows pages 1–2. Sign up to view the full content.

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%).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}