AP Biology Genetics Quiz
AP Biology Genetics Problems
Directions: Completely answer the following problems on a separate sheet of paper.
1. In 1981, a stray black cat with unusual rounded, curled black ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the "curl" cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true-breeding variety. How would you determine whether the curl allele is dominant or recessive? How would you select for true-breeding cats? How would you know they are true breeding?
I would cross my curled black ear cat with another cat of a pheno type that is known to be recessive. Depending on the offspring of the P generation I would be able to determin weather or not the allele is dominant or recessive.
To select for a true-breeding cat I would breed two cats with the curled black ears and then breed their offspring together for a few generations. If no other phenotypes emerged I would assume that I had arrived a true-breed cat.
2. B is a dominant allele coding for black fur on rabbits and b is a recessive allele coding for white fur on rabbits. Fill in the following blanks with the correct cross of the following: (1) BB x bb, (2) Bb x Bb, (3) bb x bb, (4) Bb x bb
All (100%) of the offspring are white: bb x bb
One quarter (25%) of the offspring are white: Bb x Bb
All (100%) of the offspring are black: ___ BB x bb
Three-quarters (75%) of the offspring are black: Bb x Bb
One-half (50%) of the offspring are white: Bb x bb
3. The gene for unibrow dominates over genes for split eyebrows. A unibrowed man marries a unibrowed woman. They have a split-eyebrowed child. What are the genotypes of the father, mother, and child? What is the probability of their next child having a split-unibrow?
Genotype of father is Heterozygous Bb
Genotype of Mother is Heterozygous Bb
Genotype of Child is Homozygous bb
Probability of a split-unibrow child is ¼ probability
4. Assume that attached earlobes is inherited as a dominant gene. A man with attached earlobes, whose mother had unattached earlobes and father is homozygous dominant, marries a woman with unattached earlobes. What is the probability that they will have a child with attached earlobes?
Genotype of Man is Ee
Genotype of woman is ee
Probability of offspring have attached earlobes is 1/2
5. What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs)?
AABBCC x aabbcc --- AaBbCc 100%
AABbCc x AaBbCc --- AAbbCC 1/32
AaBbCc x AaBbCc --- AaBbCc 1/16
aaBbCC x AABbcc --- AaBbCc 1/2
Assume that blood type is inherited as A and B dominant over O, but A and B incompletely dominate over each other.
6. A man whose father is type O and whose mother is type A, has a blood type A. He marries a type A woman, whose parents had the same blood types as his parents. What are the genotypes of the man and woman and what is the probability their first child will be blood type A?
Genotype of dad is iAi
Genotype of Mom is iAi
Probability of child being type A is ¾ or 75% chance
7. A type A man, whose mother was type B, marries a woman with type B blood. Their son has type B blood. This son marries a girl with type B blood. They have 6 children. 5 are type B and 1 is type O. What are the genotypes of the man, woman, son, girl, and children? Note: Some of the people may have more than one possible genotype.
Genotype of Dad is iAi heterozygote
Genotype of Mom is iBi or iBiB heterozygote or homozygote
Genotype of F1 son is iBi heterozygote
Genotype of F1 mom is iBi heterozygote
Genotype of F2 5 children with type B is iBi or iBiB heterozygote or homozygote
Genotype of F2 1 child with type O is ii homozygote
8. (dihybrid cross) Cleft chin and hairy fingers are dominant traits. A man with a cleft chin and hairy fingers (whose father had was homozygous recessive) marries a woman with a cleft chin and hairy fingers (whose father was homozygous recessive). What is the probability their first child will be without a cleft chin and no hair on its fingers?
1/16th probability
9. If the litter resulting from the mating of two short-tailed cats contains three kittens without tails, two with long tails, and six with short tails, what would be the simplest way of explaining the inheritance of tail length in these cats? Show genotypes (i.e., of all of the individuals mentioned)
Incomplete dominance inheritance pattern approximately 1:2:1 phenotypic ratio
Two short tailed cats (P) are heterozygotes (Ss)
F1 kittens without tails will be homozygous recessive (ss)
F1 kittens with short tails are heterozygotes (Ss)
F1 kittens with long tails are Homozygous dominant (SS)
10. 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 (L) |Dwarf (l) |
|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 instead of a huge Punnett square.)
homozygous for the three dominant traits 1/64th
homozygous for the three recessive traits 1/64th
heterozygous 1/8th
homozygous for axial and tall, heterozygous for seed shape 1/32nd
11. (trihybrid) In pea plants, long stems are dominant to short stems, purple flowers are dominant to white, and round seeds are dominant to wrinkled. Each trait is determined by a single, different gene. A plant that is heterozygous at all three loci is self-crossed and 2048 progeny are examined. How many of these plants would you expect to be long stemmed with white flowers, producing round seeds?
Long stemmed could be Ss (1/2) or SS (1/4) so 3/4th probability
White has to be ff or 1/4th probability
Round seeds could be Rr (1/2) or RR (1/4) so 3/4th probability
Probability of such an offspring would be 9/64th
About 288 offspring would be long stemmed, white flowered, and producing round seeds
12. Sickle cell anemia (SCA) is a human genetic disorder caused by a recessive allele. A couple plans to marry and wants to know the probability that they will have an affected child. With your knowledge of Mendelian inheritance, what can you tell them if (a) both are normal, but each has one affected parent and the other parent has no family history of SCA; and (b) the man is affected by the disorder, but the woman has no family history of SCA?
A) Both are carriers and will have 1/4th chance of producing a diseased child
B) None of the offspring will have the disease but all offspring will inevitably be carriers
13. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white. What is the simplest explanation for the inheritance of these colors in chickens? What offspring would you predict from the mating of a gray rooster and a black hen?
Incomplete dominance
50% off the offspring will be black and 50% of the offspring will be gray
14. Jimmy has type A blood. Jimmy is about to undergo a routine operation, but the doctors still want some blood on hand in case a transfusion is necessary. Jimmy’s mother has type A blood. Jimmy’s father has type AB blood. His two cousins have type O and type B.
a. List the individuals that can give blood to Jimmy. His type O cousin and his mother
b. List the individuals can’t give blood to Jimmy. His father and his type B cousin
Study Hints:
Complex Inheritance Test
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POLYGENIC PROBLEMS
When a problem says “many different genes” it is polygenic. REMEMBER:
▪ Upper case letters = Dominant
▪ Lower case letters = Recessive
▪ Look for the # of genes. For example: If height is controlled by 4 genes. . . A person who is extremely tall would have the genotype AABBCCDD.
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MULTIPLE ALLELES (BLOOD TYPE PROBLEMS)
When you are solving Blood Type problems REMEMBER multiple alleles (many ice cream flavors, only 2 scoops!) and use I’s in the Punnett Squares:
▪ Homozygous for Blood Type A or Blood Type B: IA IA , IB IB
▪ Heterozygous for Blood Type A or Blood Type B: IAi, IBi
▪ Blood Type O = ii
▪ Blood Type AB = IAIB
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SEX-LINKED PROBLEMS
When solving Sex-linked Problems REMEMBER to make an XX x XY Punnett Square first:
▪ Females = XX
▪ Males = XY
▪ The gene is ALWAYS on the X chromosome. For eg: A woman with hemophilia (h) is Xh Xh and a man with hemophilia is Xh Y (No gene on “Y”)
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MENDEL PROBLEMS
FLASHBACK! When solving problems that follow Mendel’s Pattern REMEMBER: You only need to use uppercase (dominant) or Lowercase (recessive) letters.
T t
|TT |Tt |
|Tt |tt |
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T
t
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