AP Biology Unit 5 Progress Check MCQ

Which of the following best explains how the sweet pea plants in the parental generation produce F1 offspring with 14 chromosomes?

Meiosis II and IIII lead to the formation of cells with 7 chromosomes. During meiosis II, homologous chromosomes separate. During meiosis IIII, sister chromatids separate. Two cells combine during fertilization to produce offspring with 14 chromosomes.

Meiosis produces haploid cells with half the number of chromosomes of the parent. When cells combine during fertilization, the diploid chromosome number is restored. Before meiosis, chromosomes replicate, forming duplicated chromosomes, each made of two sister chromatids. During meiosis II, homologous duplicated chromosomes line up in pairs, forming a set of 4 chromatids. The homologous chromosomes separate, forming cells with half the normal number of chromosomes, but the chromosomes are still duplicated. In meiosis IIII, the sister chromatids separate, forming cells with half the normal number of unduplicated chromosomes.

Which of the following questions would be most useful to researchers trying to determine the role of meiosis in the F2 phenotypic frequencies?

What is the recombination frequency between the genes for petal shape and pollen shape?

The F2F2 generation in this experiment showed a greater-than-expected frequency of phenotype combinations from the parental generation. The erect petals, round pollen, and hooded petals, long pollen phenotype combinations are the result of crossing over of homologous chromosomes during meiosis I. The recombination frequency in this experiment was (62+71)1118=12%(62+71)1118=12%. The closer genes are to each other, the lower the chance of them being separated from each other by crossing over and the lower the recombination frequency.

How many degrees of freedom should be used when looking up the critical value for a chi-square analysis of the ratios of phenotypes observed among the F2 offspring versus the expected phenotypic ratio assuming independent assortment?

3

The number of degrees of freedom is equal to the number of categories minus 1. In this experiment, there are 4 different phenotype combinations, and 4−1=34−1=3.

For sexually reproducing diploid parent cells, which of the following statements best explains the production of haploid cells that occurs in meiosis but not in mitosis?

Separation of chromatids occurs once, and there are two rounds of cell division in meiosis.

Separation of chromatids occurs once in meiosis, and there are two rounds of cell division, which ensures that haploid gamete cells are formed in sexually reproducing diploid organisms.

Which of the following best explains a distinction between metaphase I and metaphase II?

Homologous pairs of chromosomes are aligned during metaphase I, but individual chromosomes are aligned during metaphase II.

During metaphase II, tetrads (homologous pairs of chromosomes) are aligned along the metaphase plate. During metaphase IIII, however, single chromosomes (each composed of two chromatids) are aligned along the metaphase plate.

A compound that prevents the separation of the homologous chromosomes in anaphase I is being studied. Which of the following questions can be best answered during this study?

Is there a pattern to the movement of homologous chromosomes in the presence of this compound?

The compound prevents separation of homologous chromosomes, meaning that both the maternal and paternal chromosomes will migrate along the spindle fibers together. Patterns and other aspects of the movement of these homologous chromosomes in the presence of this compound could be recognized effectively during this study.

Based on Figure 1, which of the following questions could best be addressed?

Does synapsis of homologous chromosomes in the parent cell contribute to an increase in genetic diversity in the daughter cells?

Based on the model, synapsis of homologous chromosomes allows the opportunity for crossing over to occur, which can result in chromosomal rearrangement and increased genetic variation in the daughter cells.

Which of the following is closest to the calculated chi-square (χ2) value for the data presented in Table 1?

72.01

72.01 is the correct calculated answer using the formula χ2=Σ(o−e)2eχ2=Σ(o−e)2e.

Based on the data in Figure 1, which of the following is the best prediction of the mode of inheritance of red eyes in Japanese koi?

The allele for red eyes is inherited in an autosomal dominant pattern.

Based on the pedigree, red eye color in koi is inherited in an autosomal dominant pattern. In the second generation, two koi with red eyes produced female offspring with black eyes. This could only happen if the parents were heterozygous with the dominant allele.

Using a significance level of p=0.05, which of the following statements best completes a chi-square goodness-of-fit test for a model of independent assortment?

The calculated chi-square value is 3.91, and the critical value is 7.82. The null hypothesis cannot be rejected.

Because the null hypothesis is based on a model of independent assortment, the phenotypic ratios are expected to be 9:3:3:19:3:3:1. Using the equation χ2=Σ(o−e)2eχ2=Σ(o−e)2e, the calculated chi-square value is 3.91. With three degrees of freedom (one less than the number of phenotypic categories) and a significance level of p=0.05p=0.05, the critical value is 7.82. Because the calculated chi-square value is less than the critical value, the null hypothesis cannot be rejected. In other words, the data collected in the genetics experiment fit a model of independent assortment.

For this condition, which of the following modes of inheritance is most consistent with the observations?

Autosomal dominant

Based on the pedigree, the most likely explanation is autosomal dominant inheritance. Both parents are likely heterozygous for the condition, making it possible to have two children that are not affected. The unaffected children have a homozygous recessive genotype.

Which of the following tables best shows the expected values in the F2 generation for a chi-square goodness-of-fit test for a model of independent assortment?

Because the F1F1 offspring used to produce the F2F2 generation were heterozygous for both traits, the expected phenotypic ratio for a model of independent assortment is 9:3:3:19:3:3:1.

Which of the following best predicts the effect of adding this inhibitor to a culture of plant cells?

Plant cells will be unable to perform glycolysis due to the inhibitor and will die.

The inhibitor is a general hexokinase inhibitor, which will prevent any form of hexokinase from functioning.

The students plan to use a significance level of p=0.01. Which of the following is the most appropriate critical value for the students to use in their chi-square goodness-of-fit test?

11.34

The critical value for a significance level of p=0.01p=0.01 and 3 degrees of freedom is 11.34. A calculated chi-square value greater than 11.34 will allow the students to reject the null hypothesis of autosomal recessive inheritance.

The mean map distance between gene R and gene L is closest to which of the following?

28 map units

The recombination frequency is equal to the map distance between genes on the same chromosome. Thus, the mean recombination frequency should be determined, and this is equal to the mean map distance between the two genes.

The researchers calculated a chi-square value of 29.25. If there are three degrees of freedom and the significance level is p=0.05, which of the following statements best completes the chi-square test?

The critical value is 7.82, and the null hypothesis can be rejected because the calculated chi-square value is greater than the critical value.

The critical value is 7.82. Because the calculated chi-square value is greater than the critical value, the null hypothesis can be rejected.

Which of the follow indicates the mean number per cross of F2 plants producing medium-red grain and correctly explains the distribution of the phenotypes?

The mean number of medium-red phenotypes per cross is 20.8. The distribution of phenotypes suggests that multiple genes are involved in grain color determination.

This is the correct mean, (20+22+21+20+21)5=20.8(20+22+21+20+21)5=20.8. The phenotype distribution shows a continuous range of grain color from plants grown under the same conditions, suggesting that grain color is controlled by multiple alleles.

The mean number of fruit flies per student that are homozygous recessive for both genes is closest to which of the following?

29.0

The mean number of flies that express both recessive mutations homozygous recessive is 29.0. The sepia eyes, ebony body phenotype is homozygous recessive (eess)(eess).

Which of the following statements best explains how the fur color can be different in Himalayan rabbits raised under different temperature conditions?

The environment determines how the genotype is expressed.

The environment influences gene activity, which leads to a change in the phenotype of fur color. Temperature is the most significant environmental factor in determining the coat color of these rabbits.

Which of the following best describes an advantage of the phenotypic plasticity displayed by the tadpoles?

It gives the tadpoles increased versatility with respect to diet.

Phenotypic plasticity during development is an adaptation in Mexican spadefoot toads that allows the tadpoles to best exploit the available resources in different environments.

Which of the following best explains how the leaves from the same plant can have different stomatal densities when exposed to an elevated carbon dioxide level?

Leaf growth is promoted through increased photosynthesis, but the genetically regulated rate of stomatal production is not altered, leading to a decrease in stomatal density.

Elevated carbon dioxide levels promote leaf growth due to increased photosynthetic activity. If the number of stomata produced remains the same as the leaves grow larger, the stomatal density decreases. This is an example of phenotypic plasticity, which is an adaptive trait.

Which of the following best predicts the effect of the chromosomal segregation error shown in Figure 1?

Half of the resulting gametes will have the correct number of chromosomes, and the other half will have an incorrect number of chromosomes.

In this instance, nondisjunction of sister chromatids during meiosis IIII will produce two gametes with the correct number of chromosomes (n). The remaining two gametes that are produced will be n+1n+1 and n−1n−1, respectively.

If the woman and a man with normal clotting function have children, what is the probability of their children exhibiting hemophilia A?

50 percent for sons, 0 percent for daughters

Half of the sons would inherit the XHXH allele from the woman and a YY chromosome from the man. Without a XHXH allele, these sons would develop hemophilia AA.

Which of the following is most likely the immediate cause of the first appearance of Huntington's disease in a person?

An allele with more than 39 CAGCAG repeats was inherited by the affected person.

At some point, all alleles for the HTTHTT gene contain less than 39 CAGCAG repeats. Only after this threshold is crossed will a person inherit an allele that will cause symptoms of Huntington's disease.

(a) If a particular gene is located on the Z chromosome of this lizard species, describe why a lizard with a ZW genotype has a greater probability of expressing the recessive phenotype for the trait than a lizard with the ZZ genotype does.

(b) Using the template, construct an appropriately labeled graph, including error bars, to represent the data in Table 1. Based on the data, compare the ZZm×ZWf crosses and the ZZm×ZZf crosses to determine whether there is a significant difference between the two crosses in the fraction of female progeny produced at 24°C, 32°C, and 36°C

(c) For each cross, describe whether genetics or temperature determines the sex of progeny produced from eggs incubated at 35°C. For each cross, describe whether genetics or temperature determines the sex of progeny produced from eggs incubated at 24°C.

(d) Predict the effect of increasing global temperatures on the continued presence of the W chromosome in this species of lizard. Scientists claim that, in this species, the W chromosome is unimportant in sex determination. Instead, proteins that are encoded by a gene or genes on the Z chromosome, and that are maximally expressed only at certain temperatures, are responsible for determining whether embryos will develop as females or males. Use the data to support this claim.

If there is no dominant allele for a trait present on a homologous chromosome to mask the recessive allele, the recessive phenotype will be expressed. In this case, the gene is located on the Z chromosome. In an individual with a ZW genotype (female) there is no homologous Z partner, meaning whatever allele is present on the solitary Z chromosome will be expressed.

ZW females X ZZ males at 24°CHalf of the progeny are female. This is likely determined by genetics, as there is a 50/50 chance that the offspring inherit Z or W from mother, and always Z from fatherZZ females X ZZ males at 24°CNone of the progeny are female. This is determined by genetics and temperature, as there are no W chromosomes to pass on to the offspring, meaning all offspring will always be ZZ. At this temperature, that means no ZZ females are produced.ZW females X ZZ males at 35°CAlmost all of the progeny are female. This is likely temperature dependent, because there is equal possibility genetically that offspring could be male or female (inherit W or Z). However, the proportions are different between 24 and 36C, suggesting higher temperatures influence the sex determination. At higher temperatures, ZW and ZZ individuals develop into femalesZZ females X ZZ males at 35°CAlmost all of the progeny are female. This is likely temperature dependent, because there is equal possibility genetically that offspring could be male or female (inherit W or Z). However, at higher temperatures, ZZ offspring develop into females over males.

At temperatures over 33C, there is a sharp shift towards female sex of XX individuals rather than males. The W chromosome determines a female, but is not required to do so. Increasing temperatures can also produce females. Therefore, the presence of the W chromosome will likely reduce or even disappear entirely, and the species could just rely on temperature to determine sex.This is supported by the data showing that the presence of the W chromosome is not at all required to develop into a female. If you look at the ZZ x ZZ crosses, you can get all males or or females or a mix completely dependent on temperature, without a W chromosome in sight.

(a) Describe the process in meiosis that ensures that both maternal and paternal chromosomes are passed on to each spermatozoon.

(b) Explain why the genetic content of individual chromosomes in a spermatozoon most likely differs from the genetic content of individual chromosomes in a primary spermatocyte.

(c) In some instances, meiosis of a primary spermatocyte with six chromosomes results in two spermatozoa that contain four chromosomes and two spermatozoa that contain two chromosomes. Predict the most likely cause.

(d) A student claims that if the animal producing the spermatozoa has a mutation in a mitochondrial gene, the probability that any offspring will inherit the mutation is zero. Provide evidence to support this claim.

Meiosis is a type of cell division in which the ploidy of the organism reduces to half in the daughter cells. Meiosis produces 4 daughter cells from 1 parent cell with a different combination of the genetic material but haploid chromosomes. The meiosis process takes place in two stages in which during the first stage, the homologous chromosomes separate reducing the chromosome number to half that is from diploid chromosomes of parents to haploid in daughter. Since the process of meiosis form haploid gametes in both male and female therefore after fertilization restores the diploid chromosomes of the parent cell and form a diploid structure.

The spermatozoon is a mature cell with having haploid (N) number of chromosome while primary spermatocytes are immature cell with having diploid (2N) number of chromosomes.Spermatozoon - It is the term used to describe the motile mature sex cell of the male organism. These cells have a haploid (N) number of chromosomes. Spermatozoon has one or more swimming flagella with a head-on anterior side. This cell is ready to fertilize with the female egg cell to produce a zygote.Primary spermatocytes - This is a diploid cell having a 2N number of chromosomes. Primary spermatocyte refers to the cell before the process of Meiosis I. After Meiosis I secondary spermatocytes are produced which have half the number of chromosomes.

Nondisjunction, it occurs when chromosomes do not align and separate properly prior to the formation of daughter cells, during meiosis I or II. This means that some gametes are missing some chromosomes, and other gametes have extra chromosomes. This is usually lethal, however, some genetic disorders in humans are caused by chromosome nondisjunction. For example, Down's syndrome is caused when the original gamete has an extra copy of chromosome 21, meaning the individual contains 3 copies of chromosome 21.

Mitochondrial DNA is inherited from the maternal side. The mitochondria from the sperm cell including their DNA are degrade after fertilization.