What are the answers to the following multiple-choice genetics questions?

Here are the solutions to the multiple-choice questions: 6. Pea plants were particularly well suited for use in Mendel's breeding experiments for all of the following reasons except that: Mendel chose pea plants because they exhibit easily observable variations, allow for controlled mating, and have a relatively short generation time. An unusually long generation time would be a disadvantage for breeding experiments. The correct option is C. C) Peas have an unusually long generation time. 7. Which parental genotypes could possibly have all four blood groups expressed in their offspring? To produce offspring with all four blood groups (A, B, AB, O), both parents must carry the alleles for A, B, and O. Let's examine the cross I^Ai × I^Bi: Gametes from I^Ai: I^A, i Gametes from I^Bi: I^B, i Possible offspring genotypes: • I^AI^B (Blood type AB) • I^Ai (Blood type A) • I^Bi (Blood type B) • ii (Blood type O) This cross produces all four blood groups. The correct option is A. A) I^Ai and I^Bi 8. Why are X-linked traits more common in males than in females? Males have one X chromosome and one Y chromosome (XY), while females have two X chromosomes (XX). For an X-linked recessive trait, a male will express the trait if he inherits just one copy of the recessive allele on his X chromosome, as there is no second X chromosome to mask it. Females, however, must inherit two copies of the recessive allele (one on each X chromosome) to express the trait. The correct option is C. C) A recessive allele on the X chromosome will always produce the trait in a male. 9. Hemophilia is an X-linked recessive disorder in humans. A hemophiliac man and a non-hemophiliac woman who has hemophilic son. What is their chance of having hemophilic daughter? Step 1: Determine the genotypes of the parents. Let X^H be the normal allele and X^h be the hemophilia allele. • The hemophiliac man has genotype X^hY. • The non-hemophiliac woman has a hemophilic son (X^hY). This means she must have passed an X^h allele to her son. Since she is non-hemophiliac, her genotype must be X^HX^h (a carrier). Step 2: Perform the genetic cross between the parents. Parents: X^hY (father) × X^HX^h (mother) Possible offspring genotypes: • Daughters: X^HX^h (carrier, non-hemophiliac), X^hX^h (hemophiliac) • Sons: X^HY (non-hemophiliac), X^hY (hemophiliac) Step 3: Calculate the probability of having a hemophilic daughter. From the cross, the probability of having a daughter with genotype X^hX^h (hemophiliac) is (1)/(4) of all offspring, or (1)/(2) of all daughters. (1)/(2) = 50\% The correct option is C. C) 50% 10. Assuming that equal ratio, if a mating has already produced 3 girls, what is the probability that the next 3 children will be boys. The probability of having a boy or a girl is an independent event for each birth, approximately (1)/(2). Previous births do not influence the probability of future births. The probability of the next child being a boy is (1)/(2). The probability of the second next child being a boy is (1)/(2). The probability of the third next child being a boy is (1)/(2). The probability that the next 3 children will all be boys is the product of their individual probabilities: P(3 boys) = (1)/(2) × (1)/(2) × (1)/(2) = (1)/(8) The correct option is C. C) 1/8 11. Two true-breeding stocks of pea plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all F1 individuals have red, axial flowers. If 1,000 F2 offspring resulted from the cross, approximately how many of them would you expect to have red, terminal flowers? (Assume independent assortment) Step 1: Determine dominance and parental genotypes. • F1 generation shows all red, axial flowers, indicating that red (R) is dominant over white (r), and axial (A) is dominant over terminal (a). • True-breeding parents: Red Axial (RRAA) × White Terminal (rraa). • F1 generation genotype: RrAa. Step 2: Determine the F2 generation phenotypic ratios. The F2 generation is produced by self-crossing the F1 (RrAa × RrAa). For a dihybrid cross with independent assortment, the phenotypic ratio for two dominant traits is 9:3:3:1. • Red Axial (R_A_): (9)/(16) • Red Terminal (R_aa): (3)/(16) • White Axial (rrA_): (3)/(16) • White Terminal (rraa): (1)/(16) Step 3: Calculate the expected number of red, terminal flowers in F2. The proportion of red, terminal flowers is (3)/(16). Given 1,000 F2 offspring, the expected number is: Number = (3)/(16) × 1000 Number = (3000)/(16) Number = 187.5 Approximately 190 offspring would have red, terminal flowers. The correct option is B. B) 190 12. Protein synthesis includes which of the following two processes? Protein synthesis, also known as gene expression, involves two main stages: • Transcription: The process where DNA is used as a template to synthesize messenger RNA (mRNA). • Translation: The process where the mRNA sequence is used to synthesize a protein with the help of ribosomes and transfer RNA (tRNA). The correct option is C. C) Transcription and translation 13. If a portion of a messenger RNA molecule contains the base sequence A-A-U, the corresponding transfer RNA base sequence is During translation, transfer RNA (tRNA) molecules have anticodons that are complementary to the codons on the messenger RNA (mRNA). The base pairing rules are: • mRNA A pairs with tRNA U • mRNA U pairs with tRNA A • mRNA G pairs with tRNA C • mRNA C pairs with tRNA G Given mRNA sequence: A-A-U Corresponding tRNA anticodon sequence: U-U-A The correct option is D. D) U-U-A 14. What is the role of tRNA during translation? During translation, transfer RNA (tRNA) molecules are responsible for carrying specific amino acids to the ribosome. Each tRNA molecule has an anticodon that recognizes and binds to a complementary codon on the mRNA, ensuring that the correct amino acid is added to the growing polypeptide chain in the sequence specified by the mRNA. The correct option is A. A) Carry amino acids to the mRNA for correct placement into the protein chain *15. In a species of rabbits, there are several alleles that control for coat color. The allele B results in a brown coat color. The allele b results in a beige coat color. The two alleles are