Exam 7: Inheritance Patterns

The X and Y chromosomes are called sex chromosomes because they determine an individual's biological sex. In humans, females have two X chromosomes (XX), while males have one X and one Y chromosome (XY). These chromosomes carry genes that are responsible for the development of sexual characteristics and reproductive systems. The term "sex chromosomes" is used to differentiate them from autosomes, which are non-sex chromosomes that determine other traits and characteristics in an individual.

When Gregor Mendel crossed true-breeding pea plants with purple flowers with true-breeding plants that had white flowers, all the offspring had purple flowers because the allele for purple flowers is dominant. This means that the presence of the allele for purple flowers overrides the presence of the allele for white flowers, resulting in the expression of the purple flower phenotype in all the offspring. Since all the offspring had purple flowers, it can be concluded that the allele for purple flowers is dominant in this particular genetic cross.

Colorblindness is a sex-linked trait because the allele responsible for it is located on the X chromosome. Since males have only one X chromosome, they are more likely to be affected by colorblindness if they inherit the allele. Females, on the other hand, have two X chromosomes, so they have a lower chance of being affected. This pattern of inheritance is characteristic of sex-linked traits. Therefore, the correct answer is that colorblindness is sex-linked.

The Punnett square shown in Figure 11-1 suggests that both parents are homozygous dominant for the tall trait. This means that they both have two copies of the dominant allele for tallness. As a result, all of the offspring will inherit at least one copy of the dominant allele and will therefore be tall.

Organisms that have two identical alleles for a particular trait are said to be homozygous. This means that both alleles for that trait are the same, whether they are dominant or recessive. Homozygous individuals are genetically pure for that trait and will always pass on the same allele to their offspring. In contrast, heterozygous individuals have two different alleles for the trait, while hybrids refer to the offspring of two different species or varieties. Dominant refers to the allele that is expressed over the recessive allele in heterozygous individuals.

Hemophilia is a recessive X-linked genetic disorder, meaning it is carried on the X chromosome. Males only have one X chromosome, so if they inherit the hemophilia allele, they will express the disease. Females, on the other hand, have two X chromosomes. For a female to get hemophilia, she would need to inherit the hemophilia allele from both her mother and father. However, since most males with hemophilia do not survive to reproductive age, it is rare for a female to have a father with the disease. Therefore, it is an extremely unlikely event for a female to get hemophilia.

Blood type is controlled by multiple alleles in humans. This means that there are more than two possible variations of the gene that determines blood type. The gene responsible for blood type has three alleles: A, B, and O. Each person inherits two alleles, one from each parent, determining their blood type. The combinations of these alleles result in four blood types: A, B, AB, and O. This is why blood type is considered to be controlled by multiple alleles in humans.

In this case, the mother must be a carrier of the ALD allele. Since ALD is a sex-linked recessive trait, it means that the gene responsible for ALD is located on the X chromosome. Males have one X and one Y chromosome, while females have two X chromosomes. The father does not have ALD, so he cannot be a carrier. The elder son does not have ALD, so he cannot be a carrier either. The younger son has ALD, which means he inherited the allele from his mother. The daughter does not have ALD, but since she has two X chromosomes, she could still be a carrier. However, since the question states that the mother does not have ALD, it implies that she must be a carrier.

A female with genotype XBXb cannot be color-blind because she has one normal X chromosome (XB) which will provide the necessary gene for normal color vision. Color blindness is a recessive trait, so both X chromosomes would need to carry the color blindness allele (Xb) in order for a female to be color-blind. Since the female parent has one normal X chromosome, she cannot pass on the color blindness allele to her daughter. Therefore, a color-blind daughter cannot be born to these parents.

Sex-linked characteristics are determined by genes located on the sex chromosomes (X and Y). In humans, most sex-linked traits are located on the X chromosome. Since males have one X and one Y chromosome, they only need one copy of the gene to express the trait. In contrast, females have two X chromosomes, so they need to inherit two copies of the gene to express the trait. Therefore, sex-linked traits occur more commonly in males because they only need to inherit one copy of the gene. However, it is possible for females to express sex-linked traits if they inherit two copies of the gene or if the gene is dominant.

In a dihybrid cross, the F1 generation is produced by crossing two individuals that are heterozygous for two different traits. In this case, the genotype of the F1 individuals is AaBb x AaBB. When these individuals are crossed in the F2 generation, the expected ratio of AaBb individuals can be determined using the Punnett square. The possible genotypes of the offspring are AaBB, AaBb, AABb, and AABb. Out of these four possible genotypes, only one genotype is AaBb. Therefore, the expected ratio of AaBb individuals in the F2 generation is 1 out of 4, which can be simplified to 4/16.

A trisomy of chromosome 21 causes Down syndrome. Down syndrome is a genetic disorder that occurs when an individual has three copies of chromosome 21 instead of the usual two. This extra genetic material leads to various physical and intellectual disabilities. People with Down syndrome typically have characteristic facial features, developmental delays, and may experience other health issues such as heart defects and hearing problems.

The sex of an offspring is determined by the father because the father's sperm carries either an X or a Y chromosome. If the sperm carries an X chromosome, the offspring will be female, and if it carries a Y chromosome, the offspring will be male. The mother's egg always carries an X chromosome, so the father's contribution determines the sex of the offspring.

In this question, the genotype of the pea plant is RrYy. Each gene has two alleles, so for the R gene, there are two possible alleles (R and r), and for the Y gene, there are also two possible alleles (Y and y). To determine the number of different allele combinations in the gametes, we multiply the number of possibilities for each gene together. Therefore, the number of different allele combinations would be 2 x 2 = 4.

When a tall plant is crossed with a short plant, the resulting F1 generation will all be tall because the tall trait is dominant over the short trait. However, when these tall F1 plants are allowed to self-pollinate, their offspring (F2 generation) will show a variation in height. This is because the tall trait is determined by a dominant allele (T) and the short trait is determined by a recessive allele (t). Therefore, some of the offspring will inherit the dominant allele and be tall, while others will inherit the recessive allele and be short.

When a heterozygous tall pea plant (Tt) is crossed with a short plant (tt), the possible genotypes of the offspring are Tt and tt. Since the tall trait is dominant over the short trait, the Tt genotype will result in a tall plant. The tt genotype will result in a short plant. Therefore, out of the two possible genotypes, only one will be tall. This means that the probability of an F1 plant being tall is 50%.

Round eye shape is a recessive trait. This is because Jeanine inherited 2 alleles for round eye shape and has round eye shape, while her brother inherited 1 allele for round eye shape and 1 allele for almond eye shape and has almond eye shape. This indicates that the almond eye shape allele is dominant over the round eye shape allele, and round eye shape is only expressed when an individual has 2 copies of the recessive allele.

If a person inherits an A allele from one parent and an O allele from the other, their blood type would be type A. The A allele is dominant over the O allele, so even though they have one O allele, it will not affect their blood type.

An individual with albinism has a genotype of homozygous recessive. This means that both of their alleles for the pigmentation gene are recessive, resulting in the absence of pigmentation in their skin. Albinism is a genetic condition that is inherited when an individual receives two copies of the recessive allele for pigmentation.

The example that best illustrates Mendel's law of independent assortment is when a tall purple-flowered pea plant and a short white-flowered pea plant are crossed, producing offspring including tall white-flowered pea plants. This is because the traits of height and flower color are independently inherited, meaning that they are not linked together and can be inherited separately. In this case, the offspring inherit the tall trait from one parent and the white flower trait from the other parent, demonstrating independent assortment.

Both parents must be heterozygous because they produced both black and albino offspring. In order for the albino phenotype to be expressed, the guinea pig must have two copies of the recessive allele. Therefore, both parents must have at least one copy of the recessive allele for albinism, making them heterozygous.

A cross between a black chicken and a white chicken produces offspring with a combination of both black and white feathers, resulting in a speckled appearance. This is an example of codominance, where both alleles for feather color are expressed equally in the offspring. Incomplete dominance would result in a blending of the two colors, not the distinct speckled pattern seen here. Polygenic inheritance involves the contribution of multiple genes to a single trait, which is not the case in this scenario. Multiple alleles refers to the presence of more than two options for a specific gene, which is also not applicable here.

Keisha can identify the phenotype of the cow's hair color because the question states that the dominant allele produces brown hair in cows and the recessive allele produces white hair. The phenotype refers to the observable characteristics of an organism, in this case, the color of the cow's hair. Therefore, by observing the cow's hair color, Keisha can determine if it is brown or white, which corresponds to the phenotype. The genotype, on the other hand, refers to the genetic makeup of an organism, which cannot be determined solely by observing the cow's hair color.

Cystic fibrosis is a recessive genetic disease, meaning that both copies of the gene must have the mutation for an individual to have the disease. This means that both males and females can be affected by cystic fibrosis, as it is not limited to a specific sex. The disease is inherited in an autosomal manner, meaning it is not linked to the sex chromosomes. Therefore, the correct answer is autosomal recessive.

The genotype for Individual 1 in the diagram is Aa. This is because the disorder is determined by 2 alleles at 1 locus, and Individual 1 has one dominant allele (A) and one recessive allele (a), resulting in the heterozygous genotype Aa.

If both parents carry the recessive allele for cystic fibrosis, there is a one in four chance that their child will develop the disease. This is because both parents must pass on the recessive allele for the child to inherit the disease. In this case, there is a 25% chance that both parents will pass on the recessive allele, resulting in the child developing cystic fibrosis.

In the F2 generation, evidence of crossing-over would be apparent. This is because crossing-over is the exchange of genetic material between homologous chromosomes during meiosis, resulting in the recombination of alleles. In the F1 generation, there would be no evidence of crossing-over as the parental types would remain intact. Therefore, the correct answer is F2.

The observation that the disease is found equally in males and females suggests that it is not sex-linked, as sex-linked diseases are usually more common in one gender. Additionally, the fact that all children who had the disease had parents who also had the disease indicates that it is a dominant trait, as both parents must have at least one copy of the gene for the disease to be passed on to their children. Therefore, the gene coding for this disease is likely autosomal dominant.

The offspring being albino suggests that the allele for albino pigmentation (a) is recessive to the allele for normal pigmentation (A). This is explained by Mendel's principle of dominance, which states that one allele can mask the expression of another allele. Additionally, the fact that both parents are heterozygous for normal pigmentation (Aa) indicates that they each carry one copy of the recessive allele for albino pigmentation (a). This is explained by Mendel's principle of segregation, which states that during gamete formation, the alleles for a trait separate and each gamete receives only one allele. Therefore, the correct answer is dominance and segregation.

Based on the pedigree, it is evident that the genetic disorder is present in some of the offspring, indicating that both parents must carry at least one copy of the recessive allele. Therefore, both parents are heterozygous for the genetic disorder.

Both A and B could be true about the trait shown in the pedigree above. The pedigree shows that the trait is present in both males and females, indicating that it is not Y-linked. Additionally, the trait is passed from affected individuals to both male and female offspring, which suggests that it is not autosomal dominant. Therefore, the trait could be X-linked recessive, as it is passed from carrier mothers to affected sons, and it could also be autosomal recessive, as it is passed from carrier parents to affected offspring of both sexes.

In Drosophila fruit flies, red eyes is dominant to white eyes and the trait for eye color is found on the X chromosome. Since the female in the cross is homozygous white-eyed, she can only pass on the white-eyed allele to her offspring. On the other hand, the male is red-eyed, meaning he carries at least one red-eyed allele and can pass it on to his offspring. Since the male is not carrying any white-eyed allele, all of his offspring will have red eyes. Therefore, the proportion of the offspring expected to be white-eyed is 0%.

The arrangement of homologous chromosomes during metaphase I varies from cell to cell. This event of meiosis is important for understanding Mendel's laws of segregation and independent assortment because it is during metaphase I that the homologous chromosomes align randomly along the equator of the cell. This random alignment leads to the independent assortment of alleles, which is one of Mendel's laws. Additionally, the separation of homologous chromosomes during anaphase I, which follows metaphase I, is the mechanism that allows for the segregation of alleles, another of Mendel's laws. Therefore, the arrangement of homologous chromosomes during metaphase I is the event that reveals the mechanism for understanding Mendel's laws of segregation and independent assortment.

In the given pedigree, females 8 and 13 must have the same genotype. This can be determined by observing that both females have affected sons (squares) and unaffected daughters (circles), indicating that they are carriers of the recessive color blindness trait. Since color blindness is a sex-linked trait, it is only expressed in males. Therefore, females who are carriers will have the same genotype.

not-available-via-ai

When a pea plant that is heterozygous for round, yellow peas (RrYy) is crossed with a pea plant that is homozygous for round peas but heterozygous for yellow peas (RRYy), the offspring are expected to show two different phenotypes. This is because the traits for round peas and yellow peas are dominant, meaning that even if the offspring only inherit one copy of the dominant allele (R or Y), they will display the dominant phenotype. Therefore, the possible phenotypes for the offspring are round and yellow peas or round and green peas.