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PCB 3063 Final Exam Questions With Complete Solutions

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PCB 3063 Final Exam Questions With Complete Solutions

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PCB 3063 Final Exam Questions With
Complete Solutions

A child who is blood type A has a mother who is
blood type B. In a paternity suit, a man is accused
of being the father. He has blood type AB. Is he the
father?


A. Yes for sure

B. No for sure

C. He can not be excluded at this point.

D. There is a 50% chance he is the father. Correct
Answers C. He can not be excluded at this point.

If the mom is IBi, he could be the father.

A colorblind male would be:


X+X+

X+Y+

XcY+

XcY

,XcXc Correct Answers . XcY

A new virus has been discovered in a species of
bacteria that inhabit hot springs. The viral genome
is found to consist of four nucleotides in the
following ratios: 8% A, 42% C, 42% G, and 8% U.
The genome must be made of:


A. double-stranded DNA.

B. double-stranded RNA.

C. single-stranded DNA.

D. single-stranded RNA. Correct Answers B.
double-stranded RNA.



It is not DNA because there is U (uracil) in the
genome. Since the % of A = % of U and the % of C
= % of G, this suggests it is a double stranded RNA
genome.

A species' diploid number is 52. How many
chromosomes would be found in individuals with a
monosomy, trisomy or is a triploid?

, A. 51; 78; 53

B. 26; 53; 104

C. 78; 51; 53

D. 51; 53; 78

E. 51; 53; 104 Correct Answers D. 51; 53; 78

With 2n = 52, n = 26.

A monosomy would be 2n-1 so 52-1 = 51. A
trisomy would be 2n+1 so 52+1 = 53. A triploid
would be 3n so 3 x 26 = 78.

Can a MALE cat be tortoiseshell?

Yes

No Correct Answers Yes....IF he's XXY and a
heterozygote for the fur color genes on his X's.

Consider the following three-point cross:

ABCxabc

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