[Bio3] Guide questions for review (LT2)

Exam will take place as scheduled tomorrow, Tues. Reebop worksheets are also due then until 12nn at the Bio unit.

Coverage is Mendelian Genetics (from Mendel’s discoveries ’til problem solving). You can review by solving the prob set and questions for discussion. Presentation may be found under the Downloads tab or viewed below and answers to the Q’s for D are below.

Here are some guide questions to aid you in the concepts part. As usual, you may answer these in your notebooks for class participation points. You can consult the following sites to help you out: Biology Basics and UC Open Access.

  1. Using a Venn Diagram, compare and contrast between the following related terms:
    • gene and allele
    • character and trait
    • dominant and recessive
    • phenotype and genotype
    • genotypes of P generation and genotypes of the F1 generation
    • homozygous and heterozygous
    • law of segregation and law of independent assortment
    • law of addition and law of multiplication
  2. Enumerate some reasons why pea plants were good choices for experiments on genetics.
  3. Enumerate three of Mendel’s most significant conclusions regarding his experiments on pea plants.

Also, you may want to take a break and apply your knowledge of genetics to win at this cool (spore-like) game. Do take a screenshot of your thing.

Answers to questions in the presentation

Monohybrid cross

1. Ss x Ss
Fraction of offspring that are S_ = 3/4

2.  Yy x yy
yy offspring = 1/2
A. YY x yy
B. yy

3.
A. phenotype: all black
genotype: all Bb
B. phenotype: 3 black:1 white
genotype: 1BB:2Bb:1bb
C. phenotype: all black
genotype: 1BB:1Bb

4. WW x ww
A. phenotypic ratio: all white
genotypic ratio: all Ww
B. phenotypic ratio: 3 white: 1 yellow
genotypic ratio: 1WW:2Ww:1ww
C. phenotypic ratio: 1 white: 1 yellow
genotypic ratio: 1Ww: 1ww

Dihybrid and trihybrid cross problems
1.  RT –> tongue-roller and PTC taster
Woman’s genotype: RRtt
man genotype: RrTt
child1 genotype: RRTt
child2 genotype: RrTt
child3 genotype: Rrtt

RRtt=(1/2)(1/2)=1/4
3 of the 12 are expected to be RRtt

RRTt x RrTt
genotypes of offspring = 1RRTT:2RRTt:1RRtt:1RrTT:2RrTt:1Rrtt
phenotypes of offspring = 3 tongue-rollers, PTC tasters: 1 tongue-roller, non-PTC taster

2. A. genotypic ratio = 1AABBCC:1AABBCc:1AABbCC:1AABbCc:2AaBBCC:2AaBBCc:2AaBbCC:2AaBbCc:1aaBBCC:1aaBBCc:1aaBbCC:1aaBbCc
phenotypic ratio = 3 dominant for all three traits:1 recessive for A but dominant for B and C

B. genotypic ratio = 1AABBCC:2AABBCc:1AABBcc:1aaBBCC:2aaBBCc:1aaBBcc
phenotypic ratio = 3 dominant for all: 1 dominant for A and B, recessive for C: 3 recessive for A, dominant for B and C: 1 recessive for A, dominant for B, recessive for C

Answers to questions for discussion
1. Dick = Ffaa, Jane = Ffaa, Children’s genotype = ffaa
probability of 7th child being ffaa = 1/4

2.
A. (1) BBTT x bbtt
(2) all BbTt
(3) all black and trotting
B. (1) BBtt x bbTt
(2) 1BbTt:1Bbtt
(3) 1 black and trotting:1 black and pacing
C. (1) bbTt x Bbtt
(2) 1Bbtt:1BbTt:1bbtt:1bbTt
(3) 1 black and pacing:1 black and trotting:1 chestnut and pacing:1 chestnut and trotting
D. (1) bbtt x bbtt
(2) all bbtt
(3) all chestnut and pacing
E. (1) BbTt x BbTt
(2)1BBTT:2BBTt:1BBtt:2BbTT:4BbTt:2Bbtt:1bbTT:2bbTt:1bbtt
(3) 9 black trotting: 3 black pacing: 3 chestnut trotting:1 chestnut pacing
F.parents: B_Tt x B_Tt, 4 offspring: B_tt, parents are heterozygous for the trotting characteristic and may be either homozygous or heterozygous for the black color
G. bbtt x bbtt, probability of offspring: B_tt = 0%
H. stud = BbTt, 1st mare = B_Tt, 1st colt = B_tt
2nd mare = bbtt, 2nd colt = bbTt

3. PpRr x PpRr
A.ppR_ = 3/16
B.P_R_ = 9/16
C.pprr = 1/16
D.P_rr = 3/16
E.1/4
F.1/8
G.1/8
H.1/8

4. TtPpRr x Ttpprr
A. T_pprr = 3/16
B. ttP_R_ = 1/16
C. ttpprr = 1/16
D. T_P_R_ = 3/16
E. T_ppR_ = 3/16

5. A. phenotypic ratio = 27 two-eyed, two-horned non-flyers: 9 two-eyed, one-horned flyers: 9 two-eyed, one-horned non-flyers: 3 two-eyed, one-horned, flyers: 9 one-eyed, two-horned non-flyers: 3 one-eyed, two-horned flyers: 3 one-eyed, one-horned non-flyers: 1 one-eyed, one-horned flyers
B. 1/64 will be homozygous recessive for all three  desired traits
C. buy fresh breeding stock

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