Patterns of Inheritance
A homozygous brown mink was crossed with a silverblue mink. Brown (B) is dominant to silverblue. There were 9 offspring. What color were they? (B) Two of the offspring were mated. What would the ratio of brown to silverblue offspring be? What fraction of the offspring will be homozygous brown? Heterozygous? Homozygous recessive?
Let B represent the allele for brown color and b represent the allele for silver blue
Punnet square between homozygous brown mink and silver mink
homozygous brown mink X Silver mink
bb X BB
All the 9 offspring were brown
Mating two of the offspring
Punnet square showing f1 generation will be as follows
Bb X Bb
Ratio of brown to silverblue is 3: 1
Fraction of homozygous brown is ¼
Fraction that will be heterozygous is ½
Fraction that will be homozygous recessive is ¼
How could you determine which of the brown mink in the F2 generation in Problem 1 were homozygous and which were heterozygous?
I would perform a test cross. For brown homozygous, all the f1 generation will be brown in the test cross, whereas for brown heterozygous, 25% of the f1 generation will be silverblue
Brown homozygous test cross
BB X Bb
Brown heterozygous test cross
Bb X Bb
Yellow guinea pigs crossed with white ones always produce cream-colored offspring. Two cream guinea pigs when crossed produce yellow, cream, and white offspring in the ratio of 1:2:1. How are these colors inherited?
This is incomplete dominance, whereby offspring’s that result from cross between parents with different phenotypes, possesses a third phenotype different from that of their parent. A cross between yellow guinea pig and white guinea pig is shown in punnet square below.
Let Y be the allele for yellow color and W be the allele for white color
WW X YY
YW is the third allele cream color
A cross between two cream guinea pigs is shown below
YW X YW
It is observed that the f2 generation produces are in the phenotypic ratio of 1 yellow guinea pig: 2 cream guinea pig: 1 white guinea pig.
In horses, black is dependent upon a dominant gene (B), and chestnut upon its recessive allele (b). The trotting walk is due to a dominant gene (T), the pacing walk to its recessive allele (t). If a homozygous black pacer is mated to a chestnut homozygous trotter, what will be the phenotypic ratio of the F1 offspring?
The traits here are Coat color and Gait
The alleles for coat colors are B=Black and b=Chestnut
The alleles for gait are T=Trotting and t=Pacing
Cross between: Homozygous black pacer and chestnut homozygous trotter) BBtt X bbTT
All the f1 generation will be heterozygote black and heterozygote trotting
If two of the F1 offspring from problem 4 were mated, what would the expected phenotypic ratio be?
BbTt X BbTt
9Black and Trotting: 3 Black and Pacer: 3 Chestnut and Trotting: 1 Chestnut and Pacer
In guinea pigs R-rough, r-smooth, S-short hair, s-long hair, B-black, b-white. A rough, long haired, black male is mated to a rough, short haired, white female. After they have produced several litters, their offspring are found to be as follows: 15 rough, short haired, black; 13 rough, long haired, black; 4 smooth, short haired, black; 4 smooth, long haired, black. What were the genotypes of the parents?
This means that their father is homozygous BB.
Rough offsprings are 13+15=28.
Smooth offspring are 4+4= 8
Ratio of the two is 28/8 which is approximately 3:1. This therefore means the parents are hererozygotes, Rr
Given that long hair and short hair numbers are equal in both classes, their mother is Ss
Parental genotypes are
RrssBB and RrSsbb
In humans brown tooth enamel is inherited as a sex-linked recessive trait. A man with brown tooth enamel marries a woman with normal teeth, but whose father had brown tooth enamel. What will be the phenotypic ratio of their offspring?
Cross between brown tooth enamel man (sufferer) and normal teeth woman (carrier)
Let b represent the allele for brown tooth enamel and B normal tooth
Parental gametes XbY XBXb
Mating
F1 genotypes XBXb XBXb XBY XbY
F1 phenotypes 2 normal tooth female: 1 normal tooth boy: 1 brown tooth enamel male
Pedigree: In fruit flies, scalloped wing are dominant to normal wings. (this is NOT a sex-linked trait)
It is not possible since the normal wings are dominant to scalloped wings. The expected f1 generation is expected to have normal wings, since they are normal heterozygotes. However, this pedigree shows some offsprings with scalloped wings.
It is not possible since both parents have scalloped wings, there is no way for them to have a normal wing offspring. This pedigree shows some offsprings with normal wings.
This is possible since the parents may be heterozygous for normal wings. Some offsprings may have scalloped wings.
This is possible as in the first pedigree, there is possibility of having normal wings offspring. Also in the second pedigree it is possible since a heterozygous normal wings are crossed and there is a possibility of having scalloped wings offspring.
