What is the simplest explanation for the inheritance of these colors in chickens?
Incomplete or codominance. Feather color is controlled by 2 genes B = black and b = white. The third phenotype is the result of a 5050 mix of black and white to produce gray.
The 15 gray, 6 black, and 8 white birds represent a 2:1:1 ratio&emdash;the result of mating two heterozygous individuals: (Bb x Bb)
What offspring would you predict from the mating of a gray rooster and a black hen?
A gray rooster (Bb ) mated to a black hen (BB ) can be represented by the following Punnett square:
50% of the offspring should be gray (Bb ) and 50% black (BB )
If flower position (axial or terminal) is inherited as it is in peas what will be the ratios of genotypes and phenotypes of the generation resulting from the following cross: axialred (truebreeding) x terminalwhite?
Note: Axial (A ) is dominant over terminal (a ).
The genotypes of the parents are AARR and aarr. Therefore the gametes of the parents must be AR and ar so the genotype for all the offspring in the F1 generation will be AaRr, and their phenotype will be axialpink.
What will be the ratios in the F2 generation?
The ratio of genotypes can be determined by examining the Punnett square below:
6 axialpink  8 pink 
3 axialred  4 red 
3 axialwhite  4 white 
2 terminalpink  12 axial 
1 terminalwhite  4 terminal 
1 terminalred 
Problem 3
Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:
Character  Dominant  Recessive 
Flower position  Axial (A )  Terminal (a ) 
Stem length  Tall (T )  Dwarf (t ) 
Seed shape  Round (R )  Wrinkled (r) 
If a plant that is heterozygous for all three characters were allowed to selffertilize, what proportion of the offspring would be expected to be as follows: (Note – use the rules of probability (and show your work) instead of huge Punnett squares)
a) homozygous for the three dominant traits
AATTRR = 1/4 x 1/4 x 1/4 = 1/64
b) homozygous for the three recessive traits
aattrr = 1/4 x 1/4 x 1/4 = 1/64
c) heterozygous (assumed for each trait)
AaTtRr = 1/2 x 1/2 x 1/2 = 1/8
d) homozygous for axial and tall, heterozygous for seed shape
AATTRr = 1/4 x 1/4 x 1/2 = 1/32
What is the best explanation for this genetic situation?
Black is dominant over white
Write genotypes for the parents, gametes, and offspring.
First cross:
Parent’s genotypes  =  BB (black)  x  bb (white) 
gametes  =  B  b  
F1 offspring  =  all Bb 
Second cross:
Parent’s genotypes  =  Bb (black)  x  bb (white) 
gametes  =  B or b  b  
F1 offspring  =  Bb or bb 
There should be 50% black to 50% white offspring in this cross.
a. 318 onepod normal, 98 onepod wrinkled
Parental genotypes: PPLl x PPLl or PpLl x PPLl
b. 323 threepod normal, 106 threepod wrinkled
Parental genotypes: ppLl x ppLl
c. 401 onepod normal
Parental genotypes: PPLL x PpLL or PPLl x PPLL or PPLL x PpLl etc (nine possible genotypes).
d. 150 onepod normal, 147 onepod wrinkled, 51 threepod normal, 48 threepod wrinkled. (a 3: 3: 1: 1 ratio)
Parental genotypes: PpLl x Ppll (see below for details)
3 Onepod wrinkled (PPll , Ppll , Ppll)
1 Threepod normal (ppLl)
1 Threepod wrinked (ppll)
e. 223 onepod normal, 72 onepod wrinkled, 76 threepod normal, 27 threepod wrinkled (a 9: 3: 3: 1 ratio)
Parental genotypes: PpLl x PpLl
Father = AO (or IAi)
Mother = BO (or IBi)
First Child = OO (or ii)
What other genotypes and in what frequencies, would you expect in offspring from this marriage?
Examine the Punnett square to determine the other genotypes possible.
How many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles?
As in the ABO blood system 4 phenotypes are possible in this case:
Genotype  Phenotype 
HH, Hi  (H) 
II, Ii  (I) 
HI  (HI) 
ii  (i) 
Under these circumstances assume the following Punnett square to be true.
a. all three of their children will be of normal phenotype
3/4 x 3/4 x 3/4 = 27/64
b. one or more of the three children will have the disease (x)
1 – 27/64 = 37/64
All three have x  2 out of 3 has x  1 out of 3 has x  
+  +  =  
x x o  o o x  
3 Combinations  x o x  o x o  
o x x  x o o  
+ 3(3/4 x 1/4 x 1/4)  + 3(3/4 x 3/4 x 1/4)  = 
Note: the probability of the disease (x) = 1/4 & the probability of being normal (o) = 3/4
c. all three children will have the disease
1/4 x 1/4 x 1/4 = 1/64
d. at least one child out of three will be phenotypically normal
(Note: Remember that the probabilities of all possible outcomes always add up to 1)
1 – 1/64 = 63/64
a. aabbccdd = x x x = 1/256
b. AaBbCcDd = x x x = 1/16
c. AABBCCDD = x x x = 1/256
d. AaBBccDd = x x x = 1/64
e. AaBBCCdd = x x x = 1/128
Just remember that the probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4
How would you determine whether the curl allele is dominant or recessive?
Mate the stray to a noncurl cat. If any offspring have the “curl” trait it is likely to be dominant. If the mutation is recessive, then on ly noncurl offspring will result.
How would you select for truebreeding cats?
You know that cats are truebreeding when curl crossed with curl matings produce only curl offspring.
How would you know they are truebreeding?
A purebred “curl cat” is homozygous.
 If the trait is recessive any inividual with the “curl” condition is homozygous recessive.
 If the trait is dominant you can determine if the individual in question is true breeding (CC) or heterozygous (Cc) with a test cross (to a homozygous recessive individual).
a. AABBCC x aabbcc —> AaBbCc
b. AABbCc x AaBbCc —–> AAbbCC
c. AaBbCc x AaBbCc —–> AaBbCc
d. aaBbCC x AABbcc —> AaBbCc
2/3 x 2/3 x 1/4 = 1/9.
Genotype  
Father  AO 
Mother  BO 
OO expresses the disease
The second child’s chance of having the disease is = x = 1/16
Using the Punnett square below where P = normal pigmenation and p = white
Purple (i i P_ ) = 3
Red ( i i p p) = 1
The dominant allele I is epistatic to the p locus, and thus the F1 generation will be:
9 I_P_ : colorless
3 I_pp : colorless
3 i iP_ : purple
1 i i pp: red
Recessive
Fill in the genotypes of the individuals whose genotypes you know. What genotypes are possible for each of the other individuals?
If alkaptonuria is recessive George must be a carrier. See below.
If alkaptonuria is dominant Carla could not have the disease, as indicated in the pedigree chart, since the parents do not express the trait. See Below.
Because the daughter is normal the man’s genotype must be heterozygous for the trait so:
if X = extra digits and x = normal (5) digits then:
What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has the disease)
The probability that the baby (?) has the disease (if Elaine is a carrier) is 1/4
The total probability is 2/3 x 1/4 or 1/6.
B = Black – dominant
A = Agouti – dominant
BbAa x BbAa =

The phenotypic ratio is:
9 agouti: 4 white: 3 black
The allele is most likely dominant because the #2 individual (see below) with the trait marries a woman with the trait and 50% of their offspring are normal. If the trait were recessive one would expect the following:
100% of offspring would have the disease, which is not the case. 
What genotypes are possible for the individuals marked 1, 2, and 3?
 Bb (heterozygous)
 Bb ( if 2’s genotype were bb he would not have the disease and if BB all his children would have the condition.)
 bb (all normal individuals are homozygous recessive)