Mbchb Cat - Genetics Answer Key | MCQ Quiz | OmpathStudy Kenya

Question 1

Two pure breeding parents are crossed similar to Mendel's Parental generation. A tall plant is crossed with a short plant. What is the expected outcome for the F1 generation?

Accepted Answer

all tall — When crossing pure breeding tall (TT) with pure breeding short (tt) plants, all F1 offspring will be heterozygous (Tt). Since tall is dominant over short, all F1 plants will express the tall phenotype. ---

Question 2

If the cross from #1 is continued, what would be the expected outcome in the F2 generation?

Accepted Answer

3 tall, 1 short — F1 × F1 cross (Tt × Tt) produces: 1 TT : 2 Tt : 1 tt genotypic ratio. Since T is dominant, the phenotypic ratio is 3 tall : 1 short. ---

Question 3

Which of the following is the BEST definition of inbreeding?

Accepted Answer

mating of individuals who are closely related genetically — Inbreeding specifically refers to reproduction between genetically related individuals, which increases homozygosity and can lead to expression of deleterious recessive alleles. ---

Question 4

If a plant that has round seeds has a parent that has wrinkled seeds, what is its genotype? (Assume that round is dominant.)

Accepted Answer

but has a parent with wrinkled seeds (rr), the round-seeded plant must be heterozygous (Rr) to have inherited the recessive allele from the wrinkled parent. -- — Since the plant has round seeds (dominant phenotype) but has a parent with wrinkled seeds (rr), the round-seeded plant must be heterozygous (Rr) to have inherited the recessive allele from the wrinkled parent. ---

Question 5

Below is a Punnett square showing a cross between two parents. Use this information to respond to the next three questions. P generation: BB x bb Complete dominance: B = black rat b = white rat Referring to the Punnett square above, which of the following accurately represents the phenotypic and genotypic ratios of the F1 generation?

Accepted Answer

Phenotypic ratio 100% black, genotypic ratio 100% Bb — BB × bb cross produces all Bb offspring. Since B (black) is dominant over b (white), all offspring will be black phenotypically but heterozygous (Bb) genotypically. ---

Question 6

Using the information in the Punnett square above, how would we refer to the parents and the offspring?

Accepted Answer

One parent is homozygous dominant, one parent is homozygous recessive, the offspring are heterozygous — BB is homozygous dominant, bb is homozygous recessive, and all Bb offspring are heterozygous. ---

Question 7

If we were to cross the offspring in the Punnett square above (known as the F1 generation), what will be the genotypic and phenotypic ratios of the F2 generation?

Accepted Answer

Phenotypic ratio 3:1; genotypic ratio 1:2:1 — F1 × F1 (Bb × Bb) produces: 1 BB : 2 Bb : 1 bb (genotypic ratio 1:2:1). Phenotypically: 3 black : 1 white (ratio 3:1). ---

Question 8

Recombination frequency of < 0.5 suggests

Accepted Answer

Linkage — Recombination frequency less than 50% indicates that genes are linked on the same chromosome. Independent assortment would give 50% recombination frequency. ---

Question 9

Use the information and the Punnett square below to respond to the next three questions: Hemophilia is an X linked recessive disorder What are the chances that this couple will have a child with hemophilia?

Accepted Answer

25% — Assuming carrier female (X^H X^h) × normal male (X^H Y): 25% X^H X^H, 25% X^H X^h, 25% X^H Y, 25% X^h Y. Only X^h Y males have hemophilia = 25%. ---

Question 10

The mother undergoes amniocentesis during her pregnancy, and is told that the child she is carrying is a boy. What are the chances that her son will have hemophilia?

Accepted Answer

50% — Given the child is male, only male offspring are considered: X^H Y (normal) and X^h Y (hemophiliac) are equally likely, so 50% chance. ---

Question 11

The mother has a second pregnancy four years later. She is told that her baby will be a girl. What are the chances that her daughter will have hemophilia?

Accepted Answer

75% — Female offspring are X^H X^H or X^H X^h. Since hemophilia is X-linked recessive and father contributes X^H, no daughters can have hemophilia (they need X^h X^h). ---

Question 12

The principle that states that alleles separate during gamete formation is the

Accepted Answer

principle of segregation — Mendel's first law (principle of segregation) states that paired alleles separate during gamete formation, with each gamete receiving only one allele. ---

Question 13

If you crossed two heterozygous plants, how many of the offspring will also be heterozygous?

Accepted Answer

Half — Aa × Aa cross produces: 1 AA : 2 Aa : 1 aa. Therefore, 2/4 = 1/2 = 50% are heterozygous. ---

Question 14

A plant that has purple flowers is crossed with one that has white flowers. The offspring were half white and half purple. What were the genotypes of the parents?

Accepted Answer

Pp × pp — 1:1 ratio indicates a testcross between heterozygous (Pp) and homozygous recessive (pp) parents. ---

Question 15

Cystic fibrosis is

Accepted Answer

Autosomal recessive disorder — Cystic fibrosis is caused by mutations in the CFTR gene on chromosome 7 (autosome) and requires two copies of the defective allele. ---

Question 16

A dihybrid cross (AaBb × AaBb) will result in what offspring ratio?

Accepted Answer

9:3:3:1 — Classic dihybrid cross phenotypic ratio: 9 dominant for both traits : 3 dominant for first, recessive for second : 3 recessive for first, dominant for second : 1 recessive for both. ---

Question 17

Phenotype refers to an organism's

Accepted Answer

physical appearance — Phenotype is the observable physical or biochemical characteristics of an organism, determined by both genotype and environment. ---

Question 18

In each case where Mendel crossed true breeding plants as parents, the offspring displayed only one of the two traits seen in the parents. This observation supports which principle of genetics?

Accepted Answer

dominance and recessiveness — The observation that F1 offspring show only one parental trait demonstrates that one allele (dominant) masks the expression of another (recessive). ---

Question 19

A human being has autosomes and sex chromosomes

Accepted Answer

22 pairs, 1 pair — Humans have 23 pairs of chromosomes total: 22 pairs of autosomes and 1 pair of sex chromosomes (XX or XY). ---

Question 20

Somatic cells of a human have chromosomes and are called

Accepted Answer

46, diploid — Human somatic cells contain 46 chromosomes (23 pairs) and are diploid (2n), meaning they have two copies of each chromosome. ---

Question 21

A person who receives an extra chromosome could have

Accepted Answer

Down Syndrome — Down syndrome is caused by trisomy 21 (an extra copy of chromosome 21), making it a classic example of chromosomal abnormality due to extra chromosome. ---

Question 22

A selection type which works by constantly removing individuals from both ends of a phenotype distribution thus maintaining the same mean over time is

Accepted Answer

Stabilizing — Stabilizing selection favors intermediate phenotypes and removes extreme variants, maintaining the population mean and reducing variation. ---

Question 23

Which selection type fits sickle cell anemia?

Accepted Answer

Stabilizing — In malaria-endemic areas, heterozygotes (HbAS) have selective advantage over both homozygotes, representing balancing/stabilizing selection. ---

Question 24

Which selection type fits CCR5 mutant gene?

Accepted Answer

Directional — CCR5-Δ32 mutation provides HIV resistance, leading to directional selection favoring the mutant allele in populations exposed to HIV. ---

Question 25

What is an allele?

Accepted Answer

Alternate forms of genes — Alleles are different versions of the same gene that occupy the same locus on homologous chromosomes. ---

Question 26

Given a biallelic system (Aa) 500 people were sampled for a certain disease. One SNP locus was segregating for two alleles ("A" and "a") and three genotypes were found in the following proportions: A/A = 250; A/a = 100; a/a = 150. What is the frequency of the dominant allele?

Accepted Answer

0.6 — Total individuals = 500 - A alleles = (250 × 2) + (100 × 1) = 600 - Total alleles = 500 × 2 = 1000 - Frequency of A = 600/1000 = 0.6 ---

Question 27

What is the frequency of the recessive allele (from Q26 above)?

Accepted Answer

0.4 — Since p + q = 1, and p (frequency of A) = 0.6, then q (frequency of a) = 1 - 0.6 = 0.4 ---

Question 28

Phenylketonuria is an autosomal recessive disorder resulting from the defect of phenylalanine hydroxylase enzyme. 1/3000 Jews descendant carry the deleterious recessive allele. What is the frequency for the dominant allele?

Accepted Answer

0.982 — q² = 1/3000 ≈ 0.000333 - q = √0.000333 ≈ 0.018 - p = 1 - q = 1 - 0.018 = 0.982 ---

Question 29

What is the frequency for the recessive allele (from Q28 above)?

Accepted Answer

0.018 — As calculated above, q = √(1/3000) ≈ 0.018 ---

Question 30

The following are autosomal dominant disorders EXCEPT

Accepted Answer

Tay-Sachs — Tay-Sachs disease is an autosomal recessive disorder, while the others are autosomal dominant conditions. --- ## SECTION B (70 marks) - Essay Questions

Question 31

In incomplete dominance, when a red flower (RR) is crossed with a white flower (WW), the F1 generation will be

Accepted Answer

all pink — In incomplete dominance, neither allele is completely dominant. The heterozygous phenotype (RW) is a blend of both parental phenotypes, resulting in pink flowers. ---

Question 32

In codominance, if a person with blood type A (I^A I^

Accepted Answer

only type AB — All offspring will have genotype I^A I^B, which expresses as blood type AB due to codominance of both alleles. ---

Question 33

A cross between two F1 pink flowers (from incomplete dominance) will produce what ratio in the F2 generation?

Accepted Answer

1:2:1 — RW × RW produces RR:RW:WW in a 1:2:1 ratio, giving red:pink:white phenotypes in the same ratio. ---

Question 34

Which of the following best describes epistasis?

Accepted Answer

one gene masking the expression of another gene — Epistasis occurs when one gene's expression masks or modifies the expression of another gene at a different locus. ---

Question 35

In a complementation test, if two mutations complement each other, this indicates

Accepted Answer

they are in different genes — Complementation occurs when mutations are in different genes, allowing the wild-type phenotype to be restored in the heterozygote. ---

Question 36

The phenomenon where one gene affects multiple traits is called

Accepted Answer

pleiotropy — Pleiotropy occurs when a single gene influences multiple, seemingly unrelated phenotypic traits. ---

Question 37

In fruit flies, if the recombination frequency between genes A and B is 15%, how far apart are these genes?

Accepted Answer

15 map units — Recombination frequency directly equals map distance in centimorgans (map units). 1% recombination = 1 map unit. ---

Question 38

A test cross is performed by crossing an individual with unknown genotype with

Accepted Answer

a homozygous recessive individual — Test crosses use homozygous recessive individuals to reveal the genotype of the unknown individual through offspring ratios. ---

Question 39

In X-linked inheritance, affected males cannot pass the trait to

Accepted Answer

their sons — Males pass their Y chromosome to sons and X chromosome to daughters. X-linked traits cannot be passed from father to son. ---

Question 40

A woman who is a carrier for color blindness (X^C X^

Accepted Answer

marries a color blind man (X^c Y). What percentage of their daughters will be carriers? — Cross: X^C X^c × X^c Y produces daughters with genotypes X^C X^c (carrier) and X^c X^c (affected) in 1:1 ratio. ---

Question 41

Genomic imprinting refers to

Accepted Answer

differential gene expression based on parent of origin — Genomic imprinting involves epigenetic modifications that cause genes to be expressed differently depending on whether they were inherited from mother or father. ---

Question 42

Which of the following is an example of a chromosomal deletion syndrome?

Accepted Answer

Cri-du-chat syndrome — Cri-du-chat syndrome results from deletion of part of chromosome 5. The others are numerical chromosomal abnormalities. ---

Question 43

The karyotype 47,XXY represents

Accepted Answer

Klinefelter syndrome — Klinefelter syndrome is characterized by an extra X chromosome in males (47,XXY). ---

Question 44

Nondisjunction occurring during meiosis I will result in

Accepted Answer

all abnormal gametes — Meiosis I nondisjunction affects both chromatids of homologous chromosomes, making all resulting gametes abnormal (n+1 or n-1). ---

Question 45

Huntington's disease shows which inheritance pattern?

Accepted Answer

autosomal dominant — Huntington's disease is caused by a dominant allele on an autosome, requiring only one copy for expression. ---

Question 46

The coefficient of inbreeding (F) measures

Accepted Answer

the probability that two alleles are identical by descent — The inbreeding coefficient quantifies the probability that alleles at a locus are identical due to common ancestry. ---

Question 47

Genetic drift has the greatest effect in

Accepted Answer

small populations — Genetic drift's effect is inversely proportional to population size; smaller populations experience greater random changes in allele frequencies. ---

Question 48

The founder effect is a special case of

Accepted Answer

genetic drift — The founder effect is genetic drift that occurs when a small group establishes a new population, carrying only a subset of the original genetic variation. ---

Mbchb Cat - Genetics Answer Key | MCQ Quiz | OmpathStudy Kenya

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