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 50-50 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: axial-red (true-breeding) x terminal-white?
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 axial-pink.
What will be the ratios in the F2 generation?
The ratio of genotypes can be determined by examining the Punnett square below:
|6 axial-pink||8 pink|
|3 axial-red||4 red|
|3 axial-white||4 white|
|2 terminal-pink||12 axial|
|1 terminal-white||4 terminal|
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:
|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 self-fertilize, 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.
|Parent’s genotypes||=||BB (black)||x||bb (white)|
|F1 offspring||=||all Bb|
|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 one-pod normal, 98 one-pod wrinkled
Parental genotypes: PPLl x PPLl or PpLl x PPLl
b. 323 three-pod normal, 106 three-pod wrinkled
Parental genotypes: ppLl x ppLl
c. 401 one-pod normal
Parental genotypes: PPLL x PpLL or PPLl x PPLL or PPLL x PpLl etc (nine possible genotypes).
d. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled. (a 3: 3: 1: 1 ratio)
Parental genotypes: PpLl x Ppll (see below for details)
3 One-pod wrinkled (PPll , Ppll , Ppll)
1 Three-pod normal (ppLl)
1 Three-pod wrinked (ppll)
e. 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod 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:
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 non-curl cat. If any offspring have the “curl” trait it is likely to be dominant. If the mutation is recessive, then on ly non-curl offspring will result.
How would you select for true-breeding cats?
You know that cats are true-breeding when curl crossed with curl matings produce only curl offspring.
How would you know they are true-breeding?
A pure-bred “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.
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
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)