Hardest Questions On Genetics! Trivia Test

A child with AB blood type can receive blood from any blood type, including AB, A, B, and O. This is because individuals with AB blood type have both A and B antigens on their red blood cells, making them compatible with all other blood types.

Gregor Mendel used peas to perform his experiments. Mendel is known as the father of modern genetics, and he conducted extensive experiments on pea plants to study inheritance patterns. Peas are ideal for genetic research as they have easily observable traits, reproduce quickly, and can be easily manipulated for controlled breeding. Through his experiments with peas, Mendel discovered the basic principles of inheritance, including the concept of dominant and recessive traits.

When chickens have a codominant means of feather color inheritance, it means that both black (B) and white (W) alleles are dominant. When a black chicken (BB) is crossed with a white chicken (WW), the offspring will inherit one black allele and one white allele (BW). This results in a phenotype known as speckled, where the feathers have a combination of black and white patches. Therefore, the correct answer is speckled.

In the punnett square, there are two possible genotypes for the offspring: Gg and gg. The genotype Gg represents golden hair, while gg represents white hair. Out of the four possible combinations in the punnett square, three of them have the genotype Gg, which means that 3 out of 4 offspring will have golden hair.

The phenotype for the letter 'B' in the dihybrid cross provided is Wrinkled and Yellow. This can be determined by looking at the given options and identifying the combination of traits that includes both wrinkled and yellow. In this case, the only option that includes both traits is Wrinkled and Yellow. Therefore, Wrinkled and Yellow is the correct answer for the phenotype of the letter 'B'.

A Punnett square is a useful device for predicting the possible offspring of crosses between different genotypes. It is a grid that allows us to determine the potential combinations of alleles that can be inherited from each parent. By using this tool, we can visualize the different genotypes and phenotypes that may result from a particular cross. The Punnett square takes into account the principles of Mendelian genetics, such as the law of segregation and the law of independent assortment, to accurately predict the outcomes of genetic crosses.

When a homozygous tall plant (TT) is crossed with a homozygous short plant (tt), all the offspring will be heterozygous (Tt) because the tall trait is dominant over the short trait. This means that the offspring will have one allele for tallness and one allele for shortness.

The correct answer is C because it is the only option that is both round and green, as indicated by the key. Options A and B are not round, and option D is not green.

When a Red flower is crossed with a White flower in snapdragon flowers, the genotypes of the offspring will be RW. This is because the Red flower is represented by the genotype RR, and the White flower is represented by the genotype WW. When they are crossed, the offspring will inherit one allele from each parent, resulting in the genotype RW.

In a dihybrid cross, homozygous dominant means that both alleles for each trait are dominant. In option A, the genotype is RRYY, which means that both alleles for round and yellow traits are dominant. Therefore, option A is homozygous dominant for both traits in the dihybrid cross.

In pea plants, the tall trait (T) is dominant over the short trait (t). This means that if a plant is heterozygous, meaning it has one dominant allele (T) and one recessive allele (t), the dominant trait will be expressed and the plant will be tall. Therefore, the phenotype of a heterozygous plant in pea plants is tall.

Gregor Mendel is considered the "Father of genetics" because he conducted groundbreaking experiments with pea plants in the mid-19th century. Through his observations and meticulous record-keeping, Mendel discovered the basic principles of inheritance, including the concepts of dominant and recessive traits. His work laid the foundation for modern genetics and our understanding of how traits are passed from one generation to the next.

In simple dominant inheritance patterns, a homozygous tall plant (TT) crossed with a short plant (tt) will result in all heterozygous tall offspring (Tt). Since the tall trait is dominant, all offspring will exhibit the tall phenotype and none will be short. Therefore, the correct answer is 0%.

Based on the given information, the individual I1 is affected and has cystic fibrosis. This can be deduced from the statement that "The carriers are NOT filled in for you." This implies that the individual's genotype is not provided, and since cystic fibrosis is a recessive condition, the individual must be affected if they have it. Therefore, the phenotype of individual I1 is affected; they have cystic fibrosis.

The karyotype shows a healthy male because it contains 46 chromosomes, which is the normal number for a male. A healthy female would typically have 46 chromosomes as well, but the presence of a Y chromosome in this karyotype indicates that it is a male. There is no evidence of any extra or missing autosomes, so the karyotype is consistent with a healthy male.

When a heterozygous tall plant (Tt) is crossed with another heterozygous tall plant (Tt), there are four possible combinations of alleles in the offspring: TT, Tt, Tt, and tt. Out of these four combinations, three of them (TT, Tt, and Tt) result in tall plants, while only one combination (tt) results in a short plant. Therefore, 75% of the offspring will be tall.

A male hamster needs to have the genotype dd in order to produce only offspring with curly fur. This is because the curly fur trait is recessive and is represented by the lowercase d allele. The male hamster with genotype Dd carries both the dominant and recessive alleles, so there is a 50% chance of passing on the straight fur allele to the offspring. However, if the male hamster has the genotype dd, it will only pass on the recessive curly fur allele, resulting in all offspring having curly fur.

If a mother has blood type O and the baby has blood type B, the father cannot have blood type A. This is because blood type A is dominant over blood type O, so if the father had blood type A, there would be a possibility of the baby having blood type A as well. However, since the baby has blood type B, the father must have blood type B or AB. Therefore, Adam with blood type A could not be the father.

An individual with albinism has a genotype of homozygous recessive. This means that both copies of the gene responsible for pigmented skin are non-functional, resulting in the absence of pigmentation in the skin. Albinism is a recessive trait, meaning that an individual must inherit two copies of the non-functional gene (one from each parent) in order to express the condition.

The correct answer is Ff x Ff. In order for two people who do not have cystic fibrosis to have a child with cystic fibrosis, both parents must be carriers of the recessive allele for cystic fibrosis (Ff). When two carriers (Ff) have a child, there is a 25% chance that the child will inherit two copies of the recessive allele (ff), resulting in cystic fibrosis.

The correct answer is IAi x IBi. This is because the baby has blood type O, which is a recessive trait. Both IAi and IBi parents can pass on the i allele, resulting in the baby having blood type O. The other three sets of parents have at least one parent with a dominant allele (IA or IB), which would not result in a baby with blood type O.

The genotype of the unknown parent is aa because it is the only genotype that would result in the expression of the recessive trait. Since the unknown parent has a recessive trait, it must have two copies of the recessive allele, which is represented by the genotype aa.

The genotype Ssbb represents a white guinea pig that is heterozygous for fur length. This is because the dominant allele for short fur (S) is paired with the recessive allele for long fur (s), and the dominant allele for brown fur (B) is paired with the recessive allele for white fur (b). Since the guinea pig is heterozygous for fur length, it has one allele for short fur and one allele for long fur. Therefore, the correct answer is Ssbb.

Cystic fibrosis is inherited in an autosomal recessive manner. This means that both copies of the gene responsible for the disease need to be mutated in order for an individual to have the disease. Males and females are equally affected by cystic fibrosis, and they must inherit two copies of the mutated gene, one from each parent. If only one copy is inherited, the individual will be a carrier of the disease but will not show symptoms. Therefore, the correct mode of inheritance for cystic fibrosis is autosomal recessive.

If a couple who do not have sickle cell have a child with sickle cell, it means that both parents are carriers of the sickle cell trait. In this case, each parent would have one normal allele and one sickle cell allele. When they have another child, there is a 25% chance that the child will inherit two sickle cell alleles, resulting in sickle cell disease. The other possibilities are that the child inherits two normal alleles (25% chance), or one normal and one sickle cell allele, making them a carrier like their parents (50% chance). Therefore, the chances of the couple having another child with sickle cell is 25%.

In incomplete dominance, neither allele is completely dominant over the other, resulting in a blending of traits. In this case, red and white alleles result in pink flowers. When two pink flowers are crossed, the possible genotypes of the offspring are RR, RW, and WW. Out of these, only 25% will have the genotype RR, which corresponds to red flowers. Therefore, 25% of the offspring will be red.

When a female guinea pig homozygous dominant for black fur color is mated with a male homozygous for white fur color, the offspring will inherit one allele for black fur color from the mother and one allele for white fur color from the father. Since black fur color is dominant, all the offspring will have black fur color. Therefore, in a litter of eight offspring, there would probably be 8 black guinea pigs.

If one parent has PKU and the other does not, there is a 50% chance that each child they have will inherit the PKU gene. This is because PKU is an autosomal recessive disorder, meaning that both copies of the gene must be abnormal for the disorder to be present. In this case, the child inherited one abnormal gene from the mother, who has PKU, and one normal gene from the father. Therefore, there is a 50% chance that the parents will have another child who has PKU.

The white mouse producing only brown offspring when mated with a brown mouse suggests that the brown color is dominant over white. Since all the offspring are brown, it indicates that the white mouse does not carry the dominant brown allele. Therefore, it is most likely homozygous recessive, meaning it has two copies of the recessive white allele.

The probability of a man heterozygous for blood type A and a woman heterozygous for blood type B having a child with blood type AB is 25%. This is because both parents have one copy of the A allele and one copy of the B allele, and when combined, these alleles can result in the AB blood type in their child. However, there is also a 25% chance that the child will inherit two copies of the A allele (blood type A), 25% chance of inheriting two copies of the B allele (blood type B), and a 25% chance of inheriting two copies of the O allele (blood type O).

The person who is homozygous recessive for the cystic fibrosis trait will have cystic fibrosis. Homozygous recessive means that both copies of the gene responsible for cystic fibrosis are the recessive form, resulting in the expression of the disorder. Therefore, the individual will have cystic fibrosis, a genetic disorder that affects the lungs, digestive system, and other organs.

Pedigree C shows a sex-linked inheritance pattern. This can be determined by observing that the trait is only present in males and is passed on from an affected father to all of his daughters but not to his sons. This pattern is consistent with X-linked recessive inheritance, where the trait is carried on the X chromosome.

Homozygous recessive for seed shape means the offspring will have the genotype rr. Heterozygous for seed color means the offspring will have the genotype Rr. Since the question asks for the number of offspring that will have both of these genotypes, the correct answer is 2.

The given information states that the carriers are not filled in for the individual II3. Since the individual has Cystic Fibrosis (f), it means that they must have inherited the allele for Cystic Fibrosis from at least one of their parents. However, without knowing the genotype of the other parent, it is not possible to determine if the individual is homozygous (ff) or heterozygous (Ff) for Cystic Fibrosis. Therefore, the answer is Ff.

When two snapdragon flowers, one red and one white, are crossed and produce only pink flowers, it indicates that the pink flower color is a result of incomplete dominance. Incomplete dominance means that neither of the parental traits is completely dominant over the other, resulting in an intermediate phenotype. When the pink flowers self-pollinate, there are two possible outcomes: the offspring can inherit the pink flower color from one parent, resulting in pink flowers, or it can inherit the red or white flower color from the other parent, resulting in either red or white flowers. Therefore, the probability of the pink flowers self-pollinating to have a pink flowered offspring is 50%.

This answer is correct because the given information states that horses born to two palomino horses have a 50% chance of having a palomino coat. Incomplete dominance occurs when neither allele is completely dominant over the other, resulting in a phenotype that is a blend or intermediate of the two alleles. In this case, the palomino coat color is not completely dominant, as there is a chance for the offspring to have a different coat color.

The answer is 3 because the trait for wrinkled (r) and yellow (Y) is recessive, meaning that it requires two copies of the recessive allele to be expressed. Since the parents are both heterozygous (RrYy), there is a 25% chance for each offspring to be wrinkled and yellow. Therefore, out of the four possible combinations (RY, Ry, rY, ry), only one of them (ry) will result in offspring that are both wrinkled and yellow.

Pedigree B shows an autosomal recessive inheritance pattern because both parents are carriers of the trait and have unaffected children. In an autosomal recessive inheritance, the trait is only expressed when an individual inherits two copies of the recessive allele, one from each parent who is a carrier. This pattern is consistent with Pedigree B, where both parents are carriers and have unaffected children.

The mother rabbit's genotype is cch, c. This means that she carries one copy of the chinchillan allele (cch) and one copy of the regular coat color allele (c). Since the chinchillan allele is dominant over the regular allele, the mother rabbit exhibits the chinchillan coat color. The presence of the regular allele (c) also explains why there are offspring with the himilayan and albino coat colors, as these colors are determined by different alleles.

The probability of individuals III1 and III2 having another child with cystic fibrosis is 25%. This is because both individuals are carriers of the cystic fibrosis gene, indicated by the lowercase "f" in their genotypes. When two carriers have a child, there is a 25% chance that the child will inherit two copies of the gene and have cystic fibrosis.