Take The Easiest Biology Practice Questions
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The phenotypic expression of a gene at one locus alters that of a gene at 2nd locus is:
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Polygenic inheritance
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Pleiotropy
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Epistasis
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Incomplete dominance
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Explore key genetics concepts with 'Take The Easiest Biology Practice Questions'. This quiz assesses understanding of epistasis, pleiotropy, Mendelian genetics, and more, perfect for beginners or as a quick review for advanced learners.
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- 2.
The picture is an example of
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Epistasis
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Polygenic inheritance
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Pleiotropy
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Incomplete dominance
Correct Answer
A. PleiotropyExplanation
Pleiotropy refers to a phenomenon where a single gene affects multiple traits. In this picture, it is likely that a single gene is responsible for the expression of different characteristics or features. This is evident as the picture represents a situation where one gene has multiple effects, which aligns with the concept of pleiotropy.Rate this question:
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- 3.
When an crossing an organism that's homozygous recessive for a single trait with heterozygote ,what is the chance of a producing an offspring with the Homozygous recessive phenotype?
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0%
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25%
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50%
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75%
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100%
Correct Answer
A. 50%Explanation
When crossing an organism that is homozygous recessive for a single trait with a heterozygote, there is a 50% chance of producing an offspring with the homozygous recessive phenotype. This is because the heterozygote has one dominant allele and one recessive allele, and when crossed with a homozygous recessive organism, there is a 50% chance that the offspring will inherit the recessive allele from both parents, resulting in a homozygous recessive phenotype.Rate this question:
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- 4.
How many unique gametes could-be produced the through independent assortment by an individual with the genotype AaBbCCDdEE?
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4
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8
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16
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64
Correct Answer
A. 8Explanation
The individual has 2 different alleles for each gene, resulting in 2^4 = 16 different combinations. However, since the question asks for unique gametes, we need to consider that two of the genes (C and E) are homozygous (CC and EE). Therefore, the individual can produce 2^2 = 4 different combinations for these genes. For the remaining genes (A, B, and D), the individual can produce 2^1 = 2 different combinations for each gene. Multiplying these combinations together, we get 4 * 2 * 2 * 2 = 32 different combinations. However, we need to divide this number by 2, since the individual can produce the same gamete through different combinations. Therefore, the correct answer is 32/2 = 16/2 = 8 unique gametes.Rate this question:
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- 5.
Why did the F1 offspring of Mendel's classic pea cross always look like one of the two parental varieties?
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No genes interacted to produce the parental phenotype.
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Each allele affected phenotypic expression.
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The traits blended together during fertilization.
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One phenotype was completely dominant over another.
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Different genes interacted to produce the parental phenotype.
Correct Answer
A. One phenotype was completely dominant over another.Explanation
The F1 offspring of Mendel's classic pea cross always looked like one of the two parental varieties because one phenotype was completely dominant over another. This means that the dominant allele would always be expressed in the offspring, masking the presence of the recessive allele. As a result, the offspring would exhibit the same phenotype as the parent with the dominant allele.Rate this question:
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- 6.
Which of the following differentiates between independent assortment and segregation?
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The law of independent assortment requires describing two or more genes relative to one another.
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The law of segregation requires describing two or more genes relative to one another.
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The law of segregation requires having two or more generations to describe.
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The law of independent assortment is accounted for by observations of prophase I.
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The law of segregation is accounted for by anaphase of mitosis.
Correct Answer
A. The law of independent assortment requires describing two or more genes relative to one another.Explanation
The law of independent assortment refers to the random distribution of different genes during gamete formation. It states that the alleles for different traits segregate independently of each other. In other words, the inheritance of one trait does not affect the inheritance of another trait. This is in contrast to the law of segregation, which states that alleles for a single trait segregate during gamete formation. Therefore, the correct answer is that the law of independent assortment requires describing two or more genes relative to one another.Rate this question:
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- 7.
A sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which of the following genotypes is possible in a gamete from this organism?
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Tt
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Hh
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HhTt
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T
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HT
Correct Answer
A. HTExplanation
The genotype of the organism is HhTt, which means it has one dominant allele for head shape (H) and one recessive allele for tail length (T). In a gamete, only one allele from each gene pair is passed on. Therefore, it is possible for the gamete to have the combination of one allele for head shape (H) and one allele for tail length (T), resulting in the genotype HT.Rate this question:
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- 8.
Mendel accounted for the observation that traits which had disappeared in the F1 generation reappeared in the F2 generation by proposing that
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New mutations were frequently generated in the F2 progeny, "reinventing" traits that had been lost in the F1.
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The mechanism controlling the appearance of traits was different between the F1 and the F2 plants.
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Traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1.
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The traits were lost in the F1 due to dominance of the parental traits.
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Members of the F1 generation had only one allele for each trait, but members of the F2 had two alleles for each trait.
Correct Answer
A. Traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1.Explanation
Mendel proposed that traits can be dominant or recessive, and in the F1 generation, the recessive traits were obscured by the dominant ones. This means that even though the recessive traits were present in the F1 generation, they were not visible because the dominant traits were expressed. However, in the F2 generation, the recessive traits reappeared as they were able to be expressed when both alleles for the trait were present. This explanation aligns with Mendel's laws of inheritance and his observations on the inheritance patterns of traits.Rate this question:
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- 9.
Mendel's observation of the segregation of alleles in gamete formation has its basis in which of the following phases of cell division?
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Prophase I of meiosis
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Anaphase II of meiosis
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Metaphase I of meiosis
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Anaphase I of meiosis
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Anaphase of mitosis
Correct Answer
A. Anaphase I of meiosisExplanation
During anaphase I of meiosis, the homologous chromosomes separate and move towards opposite poles of the cell. This segregation of alleles, or different versions of genes, occurs during this phase. Mendel observed this segregation in his experiments with pea plants, where he observed that traits were inherited independently from each other. This observation led to the development of Mendel's laws of inheritance and the understanding of how alleles are passed on to offspring.Rate this question:
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- 10.
If doubly heterozygous SsNn cactuses were allowed to self-pollinate, the F2 would segregate in which of the following ratios?
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3 sharp-spined:1 spineless
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1 sharp-spined:2 dull-spined:1 spineless
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1 sharp-spined:1 dull-spined:1 spineless
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1 sharp-spined:1 dull-spined
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9 sharp-spined:3 dull-spined:4 spineless
Correct Answer
A. 9 sharp-spined:3 dull-spined:4 spinelessExplanation
When doubly heterozygous SsNn cactuses self-pollinate, the F2 generation would segregate in a ratio of 9 sharp-spined : 3 dull-spined : 4 spineless. This is because the sharp-spined trait is controlled by the dominant allele S, the dull-spined trait is controlled by the recessive allele s, the spineless trait is controlled by the dominant allele N, and the non-spineless trait is controlled by the recessive allele n. The ratio of 9:3:4 indicates that the sharp-spined trait is dominant over the dull-spined trait, and the spineless trait is dominant over the non-spineless trait.Rate this question:
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- 11.
A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N. Which of the following is a possible partial genotype for the son?
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IBIB
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IBIA
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Ii
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IBi
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IAIA
Correct Answer
A. IBiExplanation
The son's blood type is B negative, which means he must have the genotype IBi. The mother has blood type A positive, so she must have the genotype IAIA or IAi. Since the son inherited the B allele from his father, his genotype must be IBi.Rate this question:
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- 12.
Skin color in a certain species of fish is inherited via a single gene with four different alleles. How many different types of gametes would be possible in this system?
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1
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2
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4
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8
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16
Correct Answer
A. 4Explanation
In this system, skin color in the fish species is determined by a single gene with four different alleles. Each allele represents a different version of the gene that can be inherited. Since each gamete (sperm or egg) carries only one allele, the number of different types of gametes would be equal to the number of alleles. Therefore, there would be four different types of gametes possible in this system.Rate this question:
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- 13.
In one type cross of black × black, the results were as follows: 9/16 black 4/16 yellow 3/16 brown The genotype aabb must result in which of the following?
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Black
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Brown
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Yellow
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A lethal result
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White
Correct Answer
A. YellowExplanation
Based on the given results of the cross, the genotype aabb must result in yellow. This is because the genotype aabb represents a homozygous recessive individual for both black and brown alleles. Since the cross resulted in 9/16 black, 4/16 yellow, and 3/16 brown, it indicates that the yellow phenotype is present. Therefore, the genotype aabb must result in yellow.Rate this question:
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- 14.
The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the genotype of individual II-5?
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WW
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Ww
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Ww
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Ww or Ww
Correct Answer
A. WwExplanation
Individual II-5 is unaffected by the dominant trait, W. Since the trait is dominant, an individual with the genotype "ww" would not exhibit the trait. Therefore, the genotype of individual II-5 is ww.Rate this question:
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- 15.
If a person has the genotype GgTtLL , how many possiblities are there concerning the genotype of the person's gametes?
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3
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4
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6
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2
Correct Answer
A. 4Explanation
The genotype GgTtLL indicates that the person has two heterozygous gene pairs (Gg and Tt) and one homozygous gene pair (LL). Each heterozygous gene pair can produce two different gametes, one with the dominant allele (G or T) and one with the recessive allele (g or t). The homozygous gene pair (LL) can only produce one type of gamete (L). Therefore, the number of possibilities for the person's gametes is the product of the possibilities for each gene pair, which is 2 x 2 x 1 = 4.Rate this question:
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- 16.
When an organism has 2 identical alleles for a specific character , the organism is _________ for that character.
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Heterogenous
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Homologous
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Two answers are correct
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True Breeding
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Homozygous
Correct Answer
A. Two answers are correctExplanation
When an organism has 2 identical alleles for a specific character, it is considered homozygous for that character. Homozygous means that both alleles for a particular gene are the same. Therefore, the organism is homozygous for that character. Additionally, true breeding can also be considered correct as it refers to a homozygous organism that, when self-fertilized or crossed with another true-breeding organism, will always produce offspring with the same traits.Rate this question:
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- 17.
Suppose a woman with the genotype GgTt and the phenotype is black hair and blue eyes, and a man with the phenotype black eyes and blonde hair with him being heterozygous for the black eyes phenotype, what is the probability of them having a black eyed black haired child? G,g --> Hair Colour T,t --> Eye Colour
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1/4
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1/16
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0
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1/1
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1/15
Correct Answer
A. 1/4Explanation
The woman has the genotype GgTt, which means she has one dominant allele for black hair (G) and one recessive allele for blonde hair (g). The man is heterozygous for the black eyes phenotype, meaning he has one dominant allele for black eyes and one recessive allele for another eye color. Since black hair and black eyes are both dominant traits, there is a 1/4 probability that their child will inherit both dominant alleles and have black hair and black eyes.Rate this question:
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- 18.
Suppose that a species of birds has 2 possible feather colours : Dominant W - white hair Recessive w - pink hair and that birds with white feathers turn pink when they eat crabs. When breeding two birds of the genotype Ww in an environment with only crabs as a source of food , the ratio of (pink:white) feathered birds will be in F1:
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1:1
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3:1
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0:1
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1:0
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None
Correct Answer
A. 1:0Explanation
When breeding two birds with the genotype Ww, there are two possible genotypes that can be passed on to the offspring: WW and Ww. However, since white feathered birds turn pink when they eat crabs, only birds with the genotype WW will have white feathers in this environment. Therefore, all of the offspring will have the genotype Ww and will turn pink, resulting in a ratio of 1:0 for pink:white feathered birds in the F1 generation.Rate this question:
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Nov 16, 2023Quiz Edited by
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Dec 13, 2019Quiz Created by
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