Genetics Problems Key
1. tastes bitter is dominant; tasteless is recessive. the parents are heterozygous and thus the
dominant allele is completely dominant.
2. man is bb woman is either BB or Bb but BB is the most probable
3. 1/2 Bb; 1/2 bb; 1/2 wild-type; 1/2 black
4. P (AaBbcc) = 1/16;
P (A-B-C-) = 27/64
5. 1/4
6. .
D- = dark ; dd = albino
H- = short; hh = long
parental genotypes
parental phenotypes
dark, short X dark, short
Dd Hh x Dd Hh
dark, short X dark long
D- Hh x D- hh
dark, short X albino, short
Dd H- x dd H-
albino, short X albino, short
dd Hh x dd Hh
dark, long X dark long
Dd hh x Dd hh
dark, short X dark, short
D- Hh x D- Hh
dark, short X dark, long
Dd Hh x Dd hh
7. if twins are from two eggs, then the probability they will both be albino is 1/16; the probability
they will both be pigmented is 9/16
if the twins are from a single egg, then the probability they will both be albino is 1/4; the
probability they will both be pigmented is ¾.
8. IPpSs x PpSs
a. ¼
b. 1/4
c. 1/16
9. the solid black female is SsBb and the solid, red male is Ssbb
10. .
a.
b.
d.
2; one for leg hair and another for leg shape.
and c. Bowlegs B- knock-knees, bb; hairy H- smooth legs hh
Only five different parents participated in these matings. Give the genotypes of these
five individuals.
Mating
Parent 1
Parent 2
Offspring
bowlegs, hairy knees
bowlegs, hairy knees
¾ bowlegs, hairy knees
1
BbHh
BbHH
¼ knock-knees, hairy legs
bowlegs, smooth legs knock knees, smooth legs
½ bowlegs, smooth legs
2
Bbhh
bbhh
½ knock knees, smooth legs
bowlegs, hairy knees
knock-knees, smooth legs
¼ bowlegs, smooth legs
3
BbHh
bbhh
¼ bowlegs, hairy knees
Practice Genetics Problems
Winter 2005
1
4
bowlegs, hairy knees
BbHh
bowlegs, hairy knees
BBHh
¼ knock knees, hairy knees
¼ knock-knees, smooth legs
¾ bowlegs, hairy knees
¼ bowlegs, smooth legs
11. roan = RR' (heterozygous)
for the cross, RR' x R'R' expect
½ of the offspring (10 cows) to be white and
½ of the offspring (10 cows) to be roa
this is an example of co-dominance
12. husband must be father of first child (clue is blood type O)
lover must be the father of the second child (clue is blood type N)
paternity of third child can not be determined on the basis of blood types.
the wife's genotype is IAIO; NN; Rh+Rh-
13. a. possible genotypes
1/12 TTAA
2/12 TtAA
1/12 ttAA
2/12 TTAa
4/12 TtAa
2/12 ttAa
b. of the possible children
1/12 ttAA + 2/12 ttAa
3/12 or 1/4 will die before adulthood
c. of the possible children
1/12 TTAA + 2/12 TtAA
3/12 or 1/4 will be normal
d. of the possible children
2/12 TTAa + 4/12 TtAa
6/12 or 1/2 will be dwarfs without Tay-Sachs
e. of the surviving children
1/9 TTAA
2/9 TtAA
2/9 TTAa
4/9 TtAa
only 1/9 will not carry any defective alleles
14. ½ tall to ½ dwarf; ½ hairy and ½ hairless; parents TtHh and tthh (dwarf hairless)
15. Inheritance pattern: incomplete (for heterozygotes) and lethal (for homozygous recessives)
severe anemia = aa (will die at an early age); mild anemia = Aa; no anemia = AA
16. tall, yellow parent is Ttgg
dwarf, green parent is ttGg
17. gambler will pay 20$ for Ccrr and ccRr
probability
2/6 Ccrr and 1/6ccRr
if six eggs hatch and reach maturity, the gambler would be expected to pay 20$ each for 3 of them, or
$60.
poultryman will pay 30$ if CcRr and 10$ if ccrr
probability
2/6 CcRr and 1/6 ccrr
if six eggs hatch and reach maturity, the poultryman would be expected to pay 30$each for two of
them, and $10 for one of them for a total of $70.
the poultryman would be expected to pay $10 more than the gambler.
Practice Genetics Problems
Winter 2005
2