Name:        Answer Master                 .                                                               2003 03 06

 

Read the questions carefully.  Choose the best answer.  Attempt all questions, as no marks will be deducted for incorrect answers.  Value of each question is one (1) unless otherwise stated. Total 33

 

 

1.     The discoverer of the laws of genetics was            .

                A. Gregor Mendel.               C. Charles Darwin.              

                B. James Watson.                D. Francis Crick.

 

2.     The phenotype of an organism is represented by  .

A.  the symbols such as Yy

B.  the term homozygous

C.  the term heterozygous

D.  its outward appearance (the other terms refer to genotype)

 

3.     In all of Mendel’s monohybrid crosses, the F₂ plants displayed a 3:1 dominant to recessive phenotypic ratio.  Of the plants exhibiting the dominant phenotype, what proportion is homozygous?

 

A.    one-fourth

B.    one-half

C.    one-third (think of a family with 4 children – 1 son, 3 daughters. if one of the daughters is blonde, what proportion of the daughters is blonde? one of the three

D.    two-thirds 

 

4.     A heterozygote has       for the trait being studied.

 

A.    a pair of identical alleles                             C.  a haploid condition

B.    a pair of non-identical alleles                      D.   both a and c

 

5.            Of the following, which is a test cross?

 

A.    WW x WW                                                   C.  WW  x  Ww

                B.    Ww  x  ww (a test cross must include a recessive phenotype/known genotype

                D.  Ww  x  Ww

 

6.     A child with blood type O has a mother with blood type A and a father with blood type B. The parental genotypes for blood types must be          .

 

A.    I^Ai   and I^Bi  (each parent contributes one i allele to the child)

B.    I^AI^B and I^Bi

C.    I^Ai and I^BI^B

D.    I^AI^A and I^BI^B

7.     When a trait, such as height in humans, is continuously variable over a wide phenotypic range, it is reasonable to suspect that it is exhibiting            .

 

A.    epistasis

B.    multigenic inheritance (multiple genes help contribute to the spectrum of possible phenotypes)

C.    pleiotropic effects

D.    none of the above

 

8.     If an individual has the genotype Dd, the gametes produced would be            .

            A.  1/2 Dd and 1/2 Dd                                              C.  1/2 D and 1/2 d (the individual must segregate the alleles into the gametes)

            B.  3/4 Dd and 1/4 Dd                                              D.  3/4 D and 1/4 d

 

9.     In all conditions where a defective gene is on the X chromosome, transmission to the daughter can be    .

          

                A.  only through the mother

                B.  only through the father

                C.  either through the mother or the father (each parent contributes an X to a daughter)

                D.  only by mutation

 

10.   In mammals, the sex of an offspring is determined by the _________ - the only source of the______ chromosome. 

 

A.    male parent ... X

B.    female parent ... Y

C.    male parent ... Y

D.    female parent ... X

 

11.  In a cross between  AABbCc and aaBbCc, what is the probability that an offspring will have the following genotype:   Aabbcc?    

 

A.   1/4                                   C.  1/8

B.    1/16 (the offspring must inherit Aa, has a ¼ chance of getting bb, and a ¼ chance of getting cc. ¼ X ¼ = 1/16)                   

D.  1/2

 

12.   Crosses between certain white and red flowered plants result in all pink offspring in the F₁ generation. If two F₁ plants are crossed, what will be the expected phenotypic ratio in the F₂ generation?

                A.  all pink                                             C.  1 red : 1 white

                 B.  all red                                              

                D.  1 white : 2 pink : 1 red     ( key word F1 plants. They are RR’. When these two cross, the offspring will be 1RR:2RR’:1   R’R’)

 

 

13.   The appearance of the recessive trait in the offspring most probably indicates that            .

 

A.    one parent was homozygous dominant and the other parent was hybrid for the trait

B.    neither parent carried a recessive gene for the trait

C.    one parent was homozygous dominant and the other was homozygous recessive for the trait

D.    each parent carried at least one recessive gene for that trait.    (See your magic six for all three possibilities aaXaa; AaXaa, and AaXAa)

 

14.   Why are there only two alleles for the blood type group gene normally present in any one individual?

 

A.    Each parent contributes only one allele for the ABO blood group to the offspring. (We are diploid)

B.    There are not enough nucleotides in a red blood cell to produce a third allele

C.    Each allele in the ABO group must be either dominant or recessive.

D.    Blood group alleles are not segregated during meiosis.

 

15.   Assume that white colour is dominant over yellow colour in squash. If a heterozygous white-fruited plant is mated to a yellow-fruited plant, one would expect to produce          .

 

A.    all white offspring.

B.    1/2 white ; 1/2 yellow (the genotypes of the parents could be written as Ww and ww)

C.    all yellow offspring

D.    a 3:1 phenotypic ratio of white to yellow

                E.     a 3:1 phenotypic ratio of yellow and white

 

16.   Bomb disposal requires accurate colour vision.  Without knowing the exact genotype, whom would you prefer to dispose of a bomb?        

 

A.    the son of a man with normal vision and a colour-blind woman.

B.    the daughter of a man with normal vision and a colour-blind woman (In this case, we know the daughter will receive one normal X)

C.    the son of a colour-blind man and a woman who is a carrier.

D.    the daughter of a colour-blind man and a woman who is a carrier.

 

 

 

17.   If R is dominant to r, the offspring of the cross of RR and rr will              .

 

A.    be  homozygous.

B.    display the same phenotype as the RR parent (the offspring will be Rr, and therefore show the dominant phenotype)

C.    display the same phenotype as the rr parent.

D.    have the same genotype as the RR parent.

 

 

 

 

SHORT ANSWER

 

18.           A person's skin color is an example of his/her phenotype             .

 

19.           How many kinds of gametes could be produced by a RrYy plant?             four                         .

 

20.           A tall pea plant was crossed with a dwarf pea plant. Of the 20 plants in the next                 generation, 16       were tall and 4 were dwarf. What was the genotype of the original tall parent?  heterozygous or Tt

 

21.           Sometimes, one gene pair will interact to control the expression of a second gene (for example,                 coat color in Labrador retrievers). This gene interaction is called  __epistasis________.

 

 

LONG ANSWER  Each question is worth 4 marks.  Choose 3 to answer. (TOTAL = 12)

 

A. In radishes, two incompletely dominant genes control color and shape. Red and white radishes are homozygous, whereas the hybrid is purple. Long and round radishes are homozygous and if crossed will produce an oval hybrid.  Show the genotypes and phenotypes produced by crossing pure breeding red long radishes with white round radishes.

Legend RR=red RR’ =purple R’R’ = white

           LL=long LL’=oval L’L’ = round

Cross LLRR x L’L’R’R’       ALL offspring will be LL’RR’ which, according to our legend will be purple and oval.

 

B. In parakeets, four different colour patterns are possible. Read the following legend carefully.

 

B_C_  = green colour with black markings

 

B_cc  = blue pigment with black markings

 

bbC_  = yellow pigment with black markings

 

bbcc  =  white (no pigment, no markings)

 

Predict the phenotypes and ratios of a cross between a white parakeet and a heterozygous green parakeet.

 

The cross is BbCc  x bbcc Offspring will show a phenotypic ratio of 1green:1blue:1yellow:1white

 

 

C. In Drosophila, the allele for dachs  - short legs  - (d) is recessive to the allele for normal length legs (D). The allele for hairy body (h) is recessive to normal body (H).  Construct a cross between parents that are heterozygous for both characters and indicate the phenotypes of the parents. Give the phenotypes and expected ratios of the F₁ generation.

The parents’ genotypes are DdHh and they have normal body and normal length legs. Their offspring will show a phenotypic ratio

of  9 normal body/normal legs:3 normal body/short legs:3 hairy body/normal legs:1 hairy body/short legs. This is a “typical dihybrid”cross/

 

D. In a colony of space tigers, the allele for horizontal stripes (S^H) and the allele for vertical stripes (S^V) are codominant resulting in a plaid tiger when both alleles are present.  Both of these alleles are dominant over spots (s).  Predict the phenotypes and ratios of a cross between a heterozygous vertically striped space tiger and a heterozygous horizontally striped space tiger.

The cross is S^Hs  x S^Vs   The offspring will show a phenotypic ratio of 1 horizontal stripe:1 vertical stripe:1 plaid:1 spotted

 

E. A dominant gene, A, causes yellow color in rats. The dominant allele of another independent gene, R, produces black coat color. When the two dominants occur together (A- R-), they interact to produce gray. Rats of the genotype aarr are cream-colored. If a gray male and a yellow female, when mated, produce offspring approximately 3/8 of which are yellow, 3/8 gray, 1/8 cream, and 1/8 black, what are the genotypes of the two parents and the offspring? (Indicate which is which!)

The parents: the gray male is AaRr (because a cream offspring is seen, the gray male MUST be heterozygote)

                      the yellow female is Aarr (again, she’s yellow, but has a cream offspring)

The cross AaRr x Aarr.

6 of 16 will be A-R-  (3/8 gray)  6/16 comes from ¾ (Aa x Aa) X 2/4 (Rr x rr)

6 of 16 will be A-rr (3/8 yellow) same reasoning as above

2/16 will be aarr (1/8 cream) 2/16 comes from 1/4 (Aa x Aa) x 2/4 (Rr x rr)

2/16 will be aaRr (1/8 black) same reasoning as above

 

 

 

 

BONUS (2)

If the dominant gene K is necessary for hearing, and the dominant gene M results in deafness no matter what other genes are present, what percentage of the offspring produced by the cross kkMm x Kkmm will be deaf?

Offspring must have a K and mm in order to hear properly.

AND means multiply

What are the odds that an offspring will get a K? The odds are ½. What are the chances the offspring will get mm? (1/2)

So the chances of being able to hear are (1/2 x ½) one in four. So three in four or 75% of offspring from this cross will be deaf)