Test: Mendelian Genetics - 1 - MCAT MCQ

The observed phenotype of the offspring, which is gray fur resulting from the cross between a white-furred rabbit and a black-furred rabbit, suggests that the gene for fur color exhibits incomplete dominance. In incomplete dominance, neither allele is completely dominant over the other, and the heterozygous phenotype is an intermediate or blended expression of both alleles.

In this case, the white fur allele and the black fur allele are not exhibiting complete dominance, where one allele would completely mask the expression of the other. Instead, they are interacting in a way that produces an intermediate phenotype (gray fur) in the heterozygous offspring.

Mosaicism refers to the presence of cells with different genotypes in an organism, typically arising from somatic mutations during development and leading to phenotypic variation within tissues.

Codominance occurs when both alleles of a gene are expressed simultaneously and visibly in the phenotype, such as the A and B alleles in the ABO blood group system.

Complete dominance occurs when one allele is fully dominant over the other, resulting in the complete expression of the dominant allele's phenotype and the masking of the recessive allele's phenotype.

When a person with type-B blood (genotype BO) has children with a person with type-AB blood (genotype AB), their children can inherit the following blood types:

• Type-AB: The children can inherit the A and B alleles from the type-AB parent, resulting in type-AB blood.
• Type-A: The children can inherit the A allele from the type-AB parent or the B allele from the type-B parent, resulting in type-A blood.
• Type-B: The children can inherit the B allele from the type-B parent, resulting in type-B blood.

Therefore, the possible blood types for their children are type-AB, type-A, and type-B. 

Codominance occurs when both alleles of a gene are expressed equally in the phenotype of a heterozygous individual. In the case of the corn kernels, the yellow allele and the white allele are both expressed in the phenotype, resulting in an approximately even mix of yellow and white kernels in the ears of corn. Neither allele is dominant over the other, and they are both fully expressed.

Therefore, option B, "Codominance," is the correct term to describe the relationship between the yellow and white alleles for corn kernel color.

he best way to determine the genotype of a barley plant with a tall stalk and wide leaves would be to perform a testcross with a barley plant that has a short stalk and thin leaves.

A testcross involves crossing the individual of unknown genotype with an individual that is homozygous recessive for both traits. In this case, the individual with a short stalk and thin leaves would be homozygous recessive for both the tall stalk and wide leaves alleles.

By performing a testcross with a barley plant that has a short stalk and thin leaves, any offspring that exhibit the recessive traits (short stalk and thin leaves) would indicate that the tested plant is heterozygous for both traits. On the other hand, if all the offspring have the dominant traits (tall stalk and wide leaves), it would suggest that the tested plant is homozygous dominant for both traits.

In a cross of AaBb x AaBb, where A and B represent alleles for two different traits, the possible genotypes of the offspring are:

• AA, Aa, aA, aa (for the A trait)
• BB, Bb, bB, bb (for the B trait)

To determine the fraction of offspring that can be expected to express one of the two dominant alleles, but not both, we need to consider the genotypes that fulfill this condition. In this case, it would be the offspring with genotypes AaBb and Aabb (expressing the dominant A allele and the dominant B allele, respectively).

Out of the total of 16 possible combinations (4 possible genotypes for each parent), there are 2 combinations that meet this criterion (AaBb and Aabb). Therefore, the fraction of the offspring that can be expected to express one of the two dominant alleles, but not both, is 2/16, which simplifies to 1/8.

Independent assortment refers to the random segregation of alleles for different genes during the formation of gametes. This process occurs when genes are located on different chromosomes or are far apart on the same chromosome. In these cases, the genes are more likely to assort independently, meaning their inheritance is not linked or dependent on each other.

The given F1 generation results of 25% PPLL, 50% PpLl, and 25% ppll suggest that the two genes did not assort independently but rather exhibited linkage. In this case, the loci of the genes are likely to be close together on the same chromosome.

When genes are located close together on the same chromosome, they tend to be inherited together as a unit due to genetic linkage. This means that certain combinations of alleles are more likely to occur together in the offspring, while other combinations are less likely or even absent. In the provided F1 generation, the absence of other possible genotypes indicates that recombination events (such as crossing over) between the linked genes did not occur or were very rare.

Since the mother is a carrier (X-Xc) and the father is unaffected (X-Y), there are four possible combinations of parental gametes: X-X, X-Xc, Y-X, and Y-Xc.

To calculate the probability of having two male children who are both color-blind, we need to consider the probability of each event occurring.

The probability of having a male child is 0.5 (50%). The probability of passing on the color-blindness allele (Xc) from the mother is 0.5 (since she is a carrier).

To calculate the probability of both events occurring, we multiply the individual probabilities:

0.5 x 0.5 = 0.25 or 25%

Therefore, the correct answer is option D, 6.25%.

If both parents have brown eyes (genotype BB), but they produce a child with blue eyes, it means that both parents must be heterozygous carriers for the blue eye allele (genotype Bb).

There is a 25% chance (1 out of 4) of having a child with the genotype bb, which corresponds to blue eyes.

Now, we need to calculate the probability of having at least one out of the next two children with blue eyes. We can calculate the probability of the complement event (both children having brown eyes) and subtract it from 1 to get the desired probability.

The probability of having both children with brown eyes is (3/4) x (3/4) = 9/16. Therefore, the probability of having at least one child with blue eyes is 1 - (9/16) = 7/16.

Converting 7/16 to a percentage, we get approximately 43.75%, which is closest to option C, 44%.

From the Punnett square, we can see that there are 4 genotypes (BbEe, BBEe, BbEE, bbEe) that correspond to a yellow coat color. Out of the 16 possible offspring, 4 have the genotype for yellow coat color.

Therefore, the fraction of the next generation that would be expected to be yellow is 4/16, which simplifies to 1/4.

Therefore, the correct option is D. 1/4.