BIO GENETICS PROBLEMS ANSWERS
Name: ______Due: ______
Answer these questions on a separate piece of paper. Be very clear and number/letter the problems as below. Please be neat in your use of capital and lowercase letters. BOX YOUR ANSWERS please, Be sure to show your work for potential partial credit. ANSWER THE QUESTION ASKED!!!
1.
a. 0%
S / sS / SS / Ss
S / SS / Ss
b. 50%
s / sS / Ss / Ss
s / ss / ss
2. a. draw the Punnett square that illustrates this:
XC / X cXC / XC XC / XC Xc
Y / XC Y / X c Y
b.
XC / X cX c / XC X c / X c Xc
Y / XC Y / X c Y
¼ Colorblind female; ¼ normal female; ¼ colorblind male; ¼ normal male
3.a. What are the genotypes of the possible children? (show all)
At / At / at / atAT / AATt / AATt / AaTt / AaTt
AT / AATt / AATt / AaTt / AaTt
At / AAtt / AAtt / Aatt / Aatt
At / AAtt / AAtt / Aatt / Aatt
4/16 AATt; 4/16 AaTt; 4/16 AAtt; 4/16 Aatt
b. What percentage of the children will be albinos? 0%
c. What percentage will be non-tasters of PTC? 50%
4.
a. What cross would you perform to produce the most red flowered plants?
R / RR / RR / RR
R / RR / RR
Self pollinate all red flowers
b.
R / Rr / Rr / Rr
r / Rr / Rr
Cross a red plant and a white plant
5.
a. Incomplete dominance
b.
B / bB / BB / Bb
B / BB / Bb
2/4 Black; 2/4 Gray
c. Incomplete dominance
6. AO; BO
A / OB / AB / BO
O / AO / OO
¼ AB; ¼ BO; ¼ AO; ¼ OO
7.
P / pP / Pp / Pp
p / Pp / Pp
¾ Normal and ¼ Has Disease so…
¾ x ¾ x ¾ = 27/64
¼ x ¼ x ¼ = 1/64
8.
You would continue to test cross generations of possible phenotypes/genotypes and monitor the results of the offspring. If the trait is dominant, it would show up in majority if not all of the offspring. If the trait is recessive, it would show up in a minority of the offspring.
9.
A.) 25%
B.) 1/8
C.) 0%
D.) 0%
E.) 0%
F.) 3/8
10. Mother Child Could not be the Father
O B O, AO, AA
AB B AA
B A O, BO, BB
11. a. XC XC or XC X c ; XCY
b. XC X c; not colorblind ; disorder is recessive and her healthy allele dominates
c. X c Y
d. Yes; X c X c
e. Males because they have only one x chromosome
f.
XC / X cXC / XC XC / XC Xc
Y / XC Y / X c Y
Genotypes: XC XC; XC Xc ; XC Y; X c Y
Phenotypes: 2/4 healthy females; ¼ healthy male; ¼ colorblind male
g.
XC / X cX c / XC X c / X c Xc
Y / XC Y / X c Y
Genotypes: XC X c; X c Xc ; XC Y; X c Y
Phenotypes: 1/4 healthy females; ¼ colorblind female; ¼ healthy male; ¼ colorblind male
12.
a. Long winged is dominant
b. Ll x Ll
c. 98 flies
13.
a. Purple and starchy
b. PpSs x PpSs
14.
a. Codominance
b. ¼ Red; 2/4 Roan; ¼ white
15.
tX R / tX R / tX R / tX RTX r / TtX R X r / TtX R X r / TtX R X r / TtX R X r
t X r / ttX R X r / ttX R X r / ttX R X r / ttX R X r
TY / TtX R Y / TtX R Y / TtX R Y / TtX R Y
tY / ttX R Y / ttX R Y / ttX R Y / ttX R Y
Genotypes: 4/16 TtX R X r ; 4/16 ttX R X r ; 4/16 TtX R Y ; 4/16 ttX R Y
Phenotypes: 4/16 red eyed female and tan; 4/16 red eyed female and ebony; 4/16 red eyed male and tan;
4/16 red eyed male and ebony
16. A. Basic Mendelian Incomplete Dominance Sex-linked
3:1 1:2:1 2/4 Healthy Female; ¼ Healthy male; ¼ sick male
Or
¼ Healthy fem; ¼ sick fem; ¼ healthy male; ¼ sick male
B.
a. ¾ normal
b. ¼ vestigial
c. 3:1
d. If sex-linked à 100% normal females 100% vestigial males – there would be no normal males
e. Basic mendelian inheritance