ANSWERS to Part A: General Questions and Complete Dominance Problems
1. Mendel produced purebred plants by allowing plants to self-pollinate. If any offspring bred true (showed the same
trait as the parent), they were allowed to self-pollinate. After many generations of self-pollinating, if only a single
trait ever showed in the offspring, then Mendel could be sure that the plants were pure breeding.
2. Cross-pollination involves the pollen of one plant fertilizing the eggs of another plant (as opposed to self-pollination,
where plants fertilize themselves). Mendel made sure cross pollination occurred by removing the male sex organs
of flowers before they released pollen. In this way, the plants could not self-pollinate and cross-pollination
becomes the only way to produce offspring. See Figure 6.4 on page 155 in your textbook.
3. It was critical that Mendel use pure breeding plants so that he could follow with certainty the traits of interest from
one generation to the next. This allowed Mendel to make conclusions based on his data.
4. A recessive trait is the trait that is not expressed in the heterozygous individual. In order for the recessive trait to
be expressed (seen), the individual must be homozygous for that trait.
5. There are many ways to state Mendel’s First Law, The Law of Segregation. One possibility (in modern terms) is:
Hereditary characteristics are determined by distinct genes that occur in pairs. When reproductive cells are
produced, the two genes of each pair separate into different gametes. Because of this, each parent can only
contribute one gene of each pair to their offspring. See page 157 in your textbook.
6. Phenotypic ratio = 1 tall : 1 short
Genotypic ratio = 1 Tt :1 tt
7. Phenotypes of F1 = 100% white
Genotypes of F1 = 100% Ww
Phenotypic ratio of F2 = 3 white : 1 yellow
Phenotypic ratio of backcross = 2 white : 2 yellow (a 1:1 ratio)
8. If white is recessive, it is impossible to get black offspring, therefore, white must be dominant.
Assume B = white and b = black.
Three possibilities: BB x BB (no alleles for black, therefore impossible)
Bb x BB (all appear white)
Bb x Bb (possibility for white and black offspring)
Therefore, parents are both heterozygous white, Bb.
The probability that the next offspring will be black is 25% (or a 1 in 4 chance).
9. R = red and r = yellow
Two possibilities: RR x rr OR Rr x rr
In the first case, all offspring will be Rr (red). In the second case, 50% will be red (Rr) and 50% will be yellow (rr).
Therefore, the parents genotypes are Rr and rr.
10. R = purple and r = white
Purple flowered plant that is self fertilized yields a 3:1 ratio of purple:white. Therefore, the genotype of the parent
must be Rr.
R
R
RR
r
Rr
r
Rr
Only 1/3 (33%) of the purple flowered progeny will breed true
(will produce only purple progeny)
rr
11. Hybrid must be crossed with a dwarf plant….Tt x tt
T
t
t
Tt
tt
t
Tt
tt
50% offspring are tall and 50% offspring are dwarf.
12. To determine whether a black guinea pig is homozygous or heterozygous, cross it with a white guinea pig.
B = black and b = white
Possibilities…
BB x bb
B
b
B
BB
Bb
B
BB
Bb
In this case, all offspring
are black
Bb x bb
b
b
B
Bb Bb
b
bb
bb
In this case 50% of the
offspring are white.
If, after you cross a black guinea pig with a white guinea pig and you ONLY ever see black offspring, then the black
guinea pig must be BB. If you SOMETIMES get white offspring, then the black guinea pig must be Bb.
13. If a black sheep is born from white sheep, both parents must be Bb and Bb (heterozygous white). In this case, if
the farmer wants to eliminate black sheep from his flock, then he must remove the parents and the offspring.
The farmer would have to test cross all of his white sheep to with a black sheep to see if any are heterozygous
white. The farmer’s goal is to only have homozygous white (BB…no genes for black – no b allele). This is similar
to question 12.
BB x bb
B
b
B
BB
Bb
B
BB
Bb
If all offspring are white,
keep the parents.
They are BB.
Bb x bb
b
b
B
Bb Bb
b
bb
bb
If any offspring are black,
remove the parent from the flock.
They are Bb.
14. H = hornless (polled) and h = horns
The hornless bull can be either HH or Hh. The bull must be Hh in order to give an h allele to offspring (calf a).
Horned cow A must be hh since she expresses the recessive trait. Calf a must be hh since it also expresses the
recessive trait. Horned cow B must be hh since she expresses the recessive trait. Calf b must be Hh since it is
polled (received H allele from dad), but mom can only pass on recessive allele (h). Polled cow C must be Hh,
since it produces a horned calf c (hh). In order for calf c to be homozygous recessive, it must get one recessive
allele from dad and one recessive allele from mom (therefore, both parents, bull and cow C must be heterozygous
dominant).
15. Long has to be dominant to have a heterozygous long-winged fly. Long is symbolized as L.
Ll and LL
L
L
LL
l
Ll
L
LL
Ll
50% of the offspring would be homozygous long-winged.
16. G = grey and g = white
GG x gg
g
G
Gg
G
Gg
g
Gg
Gg
Phenotypic ratio = 100% grey
Genotypic ratio = 100% Gg
It would make no difference the phenotypic traits reversed genders. Each parent donates only one allele for any
trait. It doesn’t matter if it comes from mom and dad. The Punnett squares will result in the same ratio.
17. R = rough coat and r = smooth coat
a. rough x smooth  offspring, approximately ½ rough and ½ smooth
Possibilities:
RR x rr  all offspring will be Rr (rough)
Rr x rr  based on the Punnett square, 50% offspring will be rough and 50% will be smooth.
Therefore, the parent genotypes must be Rr and rr. Their offspring will either be Rr (if they are rough) or rr (if
they are smooth).
b. Rr (rough parent) x Rr (rough F1)
R
r
R
RR
Rr
r
Rr
rr
Phenotypic ratio = 3 rough : 1 smooth
Genotypic ratio = 1 RR : 2 Rr : 1 rr
18. B = brown and b = blue
Possibilities:
Bb x bb  based on Punnett square, 50% will have brown eyes (Bb) and 50% will have blue eyes (bb)
BB x bb  based on Punnett square, all children will have brown eyes (Bb).
Therefore, brown-eyed father must be Bb and blue-eyed mother must be bb.
19. grey must be dominant…GG = grey, Gg = grey and gg = white
Parents
Female x Male
grey x white
grey x grey
grey x white
grey x grey
Progeny
Female Parent Genotype
Grey
White
81
82
Gg
Approximately 50% grey & 50% white…if male is gg, then female must be Gg (to
have white offspring, one g allele comes from dad, and the other g allele must
come from mom)
118
39
Gg
Approximately 75% grey % 25% white (3:1 ratio)…therefore, both parents must be
heterozygous grey (Gg x Gg)
74
0
GG
100% off offspring are grey. If male is gg, then female must be homozygous grey,
GG, to ensure 100% offspring are grey. If she were Gg, then 50% would be grey
and 50% would be white.
90
0
GG or Gg
100% off offspring are grey. Three possibilities for parents…GG x GG (mom x
dad) or Gg x GG (mom x dad) or GG x Gg (mom x dad). Both parents cannot be
Gg and Gg, otherwise you will get 3 grey to 1 white.
20. A = normal and a = albino
a. Aa (a carrier for albinism) x aa  based on the Punnett square, 50% will be normally pigmented (but will be a
carrier) and 50% will be albino.
b. The chance that any pregnancy would result in an albino child is 2/4 or 1/2 (50%).
21. N = normal and n = cystic fibrosis (CF)
In order to be normal, BUT have children with CF, both parents MUST be Nn.
a. Based on Punnett square between Nn and Nn, there is a 25% chance (or a 1 in 4 chance) that their next child
will have CF.
b. From the same Punnett square used in part (a), it is found that there are 3 non-affected children out of 4
possibilities. Two of these three non-affected children are carriers. Therefore, there is a 67% chance (or 2 out
of 3 chance) that any given child is a carrier for CF.