life science

  1. At the initial concentration, side B(4.0 M glucose) relative to A(2.0 M glucose), is:



    B. There is a higher concentration of glucose on side b. To reach equilibrium, water will move from side A to side B. Thus, initially side B is hypertonic to side A
  2. Which molecule moves across the membrane in order to reach equilibrium?



    C. Glucose does not diffuse across a semi-permeable membrane. However, the smaller water molecules will diffuse across the membrane by osmosis from side A to side B until both sides are isotonic (equal)
  3. F1 father A A mother aa
    F2 father A a mother Aa
    A normal (AA) male is mated with an allomentic (aa) female. What is the percentage of offspring in the F1 generation (first generation after crossing two parental lines) with the disease:
    a. 100%
    b. 75%
    c. 25%
    d. 0%
    d. The disease, Allomentia is carried on the gene "a" which occurs in all progency of the F1 generation in a heterozygous Aa state. This dominant gene "A" wil suppress phenotypic expression of the disease in the entire generation
  4. The percentage of F2 offspring (generation produced from F1 generation) that are carriers of allomentia is:



    D. 50%
  5. The percentage of F2 offspring (generation produced from F1 generation) that are carriers of allomentia is:



    C. Note that each box of the Punnett swuar in F1 and F2 represents 25% of the generation. When the % is converted to a decimal and the square root extracted, the individual gene frequency is derived. Also, the allele frequency canbe determined by counting the number of each allele (A,a) and dividing the total to determne the proportion
  6. When a normal (XY) male mates witha color-blinded (xx) female, the percentage of color-blinded children (male:female) in the F1 generation will be:



    A. All the F1 males are color-blind because the gene carried on the X chromosome is not recessive to the Y. This gene is recessive to the X in females, which makes all F females carriers
  7. Color-blindness is a recessive trait found on the X chromosomes. If a color blind mother mates with a normal father, the mother can give birth to:



    C. Because females possess 2 X chromosomes with one dominant to sex-linked genes, a color-blind mother can ony produce carrier daughters and color-blind sons when the father is normal
  8. UUA - GCG - AUA - CGC mRNA (codon)
    CGC - GCG - UAU - AAU tRNA (anticodon)
    aa1 aa2 aa3 a4 Amino Acid (aa)
    Translate the mRNA (codon) from left to right by selecting the proper amino acid sequence to create the corresponding tRNA (anticodon)



    • A. The base pairing on the mRNA must be compatibly matched with those of the tRNA as follows:
    • UUA GCG AUA CGC
    • AAU CGC UAU GCG
    • 4 1 3 2
  9. Which statement is correct?



    D. RNA contains the sugar ribose
  10. If the alleles of the parents are known, how does a Punnett square help in predicting the genotype of the offspring?
    A trait can be or discouraged based on the probabilities found when crossing two parents to create offspring.

    Shows what parents would be the best to cross to encourage a trait
  11. In each dilution from tube B to tube E, the number of bacteria/ml is reduced by:



    D. 9/10
  12. If the number of cells in tube D is 500/ml, the number in tube B was;



    • D. 50,000
    • The cell count in each vial is 0.1 the count in the preceding vial (e.g. vial D=500, vial C=5,000, vial B=50,000)
  13. At position K on the bacterial growth curve, the expression ^N/^ is close to :



    B. 0.0
  14. Choose the correct statement.
    a. Maximum groth occurs at K
    b. Low growth occurs at D
    c. Maximum N/t occurs at L
    d. Minimum groth occurs at L
Author
anh
ID
79750
Card Set
life science
Description
life science
Updated