A question from s3jealousy: Genetics Problems w/ PUNNET SQUARES?
Please I am so lost with this.
1. In humans, blue eyes are recessive to brown eyes. Draw a punnet square showing the expected ratios of chilren born to a blue-eyed woman and a brown-eyed man whose mother was blue eyed.
2. In cats, short hair is dominant to long hair. A shorthaired tomcat is mated to a longhaired female. She has eight kittens: 6 shorthaired and 2 longhaired. Diagram the cross. How do the results compare to the expected ratios?
3. Cystic fibrosis is recessively inherited trait. Two normal people have a child who has cystic fibrosis. What was the probability that this couple produces a child with cystic fibrosis. What is the probablility that this couple will have two children in a row afflicted with cystic fibrosis?
4. The gene for red/green eye color vision is sex-linked. Red/green color blindness is recessive. Would the children of a color-blind woman and a man of normal color vision be color-blind?
5. The chestnut color of horses is due to a recessive gene, while the dominant allele results in a black coat. In terms of gaits, a horse may be a trotter (dominant) or a pacer (recessive). By means of a Punnet Square, show the offspring which could result from a cross between a black trotter whose mother was a chestnut pacer and a black pacer whose father was chestnut.
6. Listed below are the progeny from two crosses between a species of flowering plants. What pattern of inheritance does this cross illustrate? Determine the genotypes of the parents and their progeny. Draw a punnet square to validate your answers. Show all work.
– Cross 1: blue-flowered plant X white-flowered plant gives F1: all pale-blue flowered
– Cross 2: pale-blue F1 X pale-blue F1 gives F2: 27 blue, 49 pale-blue, and 24-white flowered plants.
Most detailed answer:
Answer by Dina Felice
1. bb (woman) x Bb (man)
___ B | b
b | Bb | bb
2. Ss x ss
___ S | s
s | Ss | ss
We would expect 50-50, so 75-25 is unexpected. It is as if the female was also heterozygous, instead of homozygous long.
3. 25% or 1/4.
6.25% or 1/16
4. The sons would all be colorblind, none of the daughters would be, but all the daughters would be carriers.
5. BbTt x Bbtt
You would see 4 possible phenotypes, black trotters, black pacers, chestnut trotters and chestnut pacers, but it is 32 squares, so I’m not going to write it out.
6. This is incomplete dominant. You are crossing 2 true breeding flowers, then getting all heterozygous offspring, then normal 1:2:1 ratio in F2.
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