Presentation on theme: "1. A form of a gene is called a(n) _________. (For example, the gene for flower color in pea plants has two forms, one for purple and one for white. These."— Presentation transcript:
1. A form of a gene is called a(n) _________. (For example, the gene for flower color in pea plants has two forms, one for purple and one for white. These are possible ________ for this gene.) A.Genotypes B.Genomes C.Alleles D.Proteomes
2. A recessive allele is A.a form of a gene that is always expressed B.a form of a gene that can be hidden C.the allele that is less common in the population D.both B and C
3. Describing the physical appearance/ expression of a trait defines the _______________ of the individual. A.Phenotype B.Genotype C.Prototype D.Gene pool
4. Identifying the two alleles possessed by an individual for a trait (such as Rr) defines their __________________. A.Phenotype B.Genotype C.Prototype D.Gene Pool
5. An organism with two different alleles for a trait such as (Gg) is called A.Homozygous B.Heterozygous C.Haploid D.Diploid
The prefix hetero= different. A heterozygous genotype is one where the individual has two different alleles (forms of the gene) for a trait.
6. Which of the following is true regarding the genotype of an individual with a dominant phenotype for a trait? A.It must be homozygous B.It must be heterozygous C.It could be homozygous or heterozygous
7. Which of the following is true regarding the genotype of an individual with the recessive phenotype for a trait. A.It must be homozygous B.It must be heterozygous C.It could be homozygous or heterozygous
The black fur allele is dominant in rabbits. Rabbits will have dark fur whether they have one black allele or two. (Homozygous dominant or heterozygous will look the same.) Rabbits can only have white fur if both of their alleles are the recessive white fur allele. (Must be homozygous recessive.)
8. Which of the following is the correct way to set up a Punnett square for a cross of two cats that are both heterozygous for bushy tails? The bushy tail allele is B and the nonbushy tail allele is b. A.. B.. C.. Bb BB b b Bb B b
o Each side of the Punnett square represents the possible alleles that can be passed down from one parent. o Since each gamete (egg or sperm) has only one allele for each gene, only one letter should be placed over each square. o A heterozygous parent has one of each allele, so 50% of the gametes have one allele and 50% of the gametes have the other. B b B b Father’s sperm Mother’s egg
9. If the purple flower allele (P) is dominant and the white flower allele is recessive (p), can two plants with purple flowers produce a seed for a plant with white flowers? A.Yes B.No
If both are heteozygous for the trait, they may each pass down the white allele.
10. If the purple flower allele (P) is dominant and the white flower allele is recessive (p), can two plants with white flowers produce a seed for a plant with purple flowers? A.Yes B.No
Since the white flower is the recessive allele, plants with this recessive trait must have two recessive alleles, and therefore do not have any purple flower alleles to pass down. X=
11. The widow’s peak allele is dominant (W), the nonwidow’s peak allele is recessive (w). If one parent is heterozygous for the allele (Ww) and the other parent does not have a widow’s peak (ww). What is the probability that their first child has a widow’s peak? (Hint: Draw a Punnett square) A.100% B.75% C.50% D.25% E.0%
12. The parent’s from the previous problem have two children with widow’s peaks. What is the probability that their third child will have a widow’s peak? A.100% B.75% C.50% D.25% E.0%
13. Hitchhiker’s thumb is a recessive trait (t). A nonhitchhiker’s thumb is dominant (T). If two parents are heterozygous for the trait, what is the probability that they will have a child with a hitchhiker’s thumb? A.100% B.75% C.50% D.25% E.0%
Tt TTTTt t tt 1 out of 4 (25%) probability to inherit recessive alleles from both parents.
14. The leg-headed (legs instead of antennae) allele is recessive in fruit flies (a). If a homozygous dominant fruit fly for the normal phenotype (AA) is mated with a leg- headed fruit fly (aa), what is the probability that they will have leg-headed offspring? A.100% B.75% C.50% D.25% E.0% Normal fruit fly Leg-headed fruit fly (antennaepedia)
AA aAa a All the offspring will inherit a dominant allele from the homozygous dominant fruit fly. Therefore they can not have an offspring with the recessive trait.
15. A true-breeding yellow flowered plant is crossed with a true-breeding white flowered plant. The F 1 generation has all yellow flowers. What is the probably outcome for the F 2 generation? A.All yellow B.All white C.50% yellow, 50% white (1:1) D.75% yellow, 25% white (3 yellow: 1 white) E.75% white, 25% yellow (3 white: 1 yellow)
The true-breeding plants must be homozygous. (YY x yy) The F 1 generation must be heterozygous (Yy). Yellow must be dominant. The F 2 produced by the F 1 cross (Yy x Yy) should have a 3 to 1 phenotype ratio of yellow to white. Yy YYYYy y yy
16. What is the genotype ratio from a cross of two plants that are heterozygous for round seeds (R= round, r = wrinkled). A.3:1 B.1:1 C.1:2:1 D.9:3:3:1
In a heterozygous cross, 3 genotypes can be produced: RR, Rr or rr. The probabilities are: 1 out of 4 RR 2 out of 4 Rr (can inherit R from egg and r from pollen or r from egg and R from pollen) 1 out of 4 rr_________________ 1:2:1 ratio Rr RRRRr r rr Pollen Egg
17. The allele for purple flowers is dominant (P) in peas. White is recessive (p). If you have a pea plant with purple flowers which you want to determine is heterozygous or homozygous, which of the following is true? A.It must be homozygous B.It must be heterozygous C.You could perform a test cross, which is usually crossing with a recessive plant D.You could perform a test cross, which is usually crossing with a homozygous dominant plant
A test cross can be used to determine the genotype of a organism with a dominant phenotype. It involves crossing with a recessive. Recessive offspring can only be produced if the organism is heterozygous. If the results are all offspring with the dominant phenotype, the test may be inconclusive (unless there is a very large number of offspring)
1.C 2. B 3. A 4. B 5. B 6. C 7. A 8. B 9.A 10.B 11.C 12.C 13. D 14. E 15. D 16. C 17.C