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MENDEL AND THE GENE IDEA

CHAPTER 14. MENDEL AND THE GENE IDEA. Due to the arrangement of the flower petals peas had always self pollinated so they were pure or homozygous for every trait. True breeding. homozygous. Testcross. SOLVING GENETICS PROBLEMS FOR DUMMIES 1. Read the problem and underline the key words.

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MENDEL AND THE GENE IDEA

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  1. CHAPTER 14 MENDEL AND THE GENE IDEA

  2. Due to the arrangement of the flower petals peas had always self pollinated so they were pure or homozygous for every trait. True breeding

  3. homozygous

  4. Testcross

  5. SOLVING GENETICS PROBLEMS FOR DUMMIES 1. Read the problem and underline the key words. 2. Write out your key. B=black dominant; b=white recessive 3. Write out the P cross. male first then female 4. Determine the gametes that can be produced by each parent. 5. Draw a Punnett Square of proper size and write male gametes along side and female gametes across the top. 6. Fill in the Punnett Square. Always put capital letters in the pair first. 7. Write out your F1 genotypic percentages: F1g%=25% AA; 50% Aa; 25% aa 8. Write out your F1 phenotypic percentages: F1p%= 75% black 25% white

  6. Page 1 16. F1g%: 12.5% AABb 12.5% AAbb 25% AaBb 25% Aabb 12.5% aaBb 12.5% aabb 17. F1g%: 6.25% AABB 12.5% AABb 12.5% AaBB 25% AaBb 6.25% AAbb 12.5% Aabb 6.25% aaBB 12.5% aaBb 6.25% aabb

  7. Page 3 1. P: Bbnn X bbNn 2. P: BbNn X bbnn F1g%: 25% BbNn 25% bbNn 25% Bbnn 25% bbnn F1p%: 25% black, normal 25% brown, normal 25% black, short 25% brown, short F1g%: 25% BbNn 25% Bbnn 25% bbNn 25% bbnn F1p%: 25% black, normal 25% black, short 25% brown, normal 25% brown, short

  8. page 3 #3 F1 cross: BbNn X BbNn P: BBNN X bbnn F1p%: 100% Black, long furred F2g%: 6.25% BBNN 12.5% BBNn 12.5% BbNN 25% BbNn 6.25% BBnn 12.5% Bbnn 6.25% bbNN 12.5% bbNn 6.25% bbnn

  9. page 3 #3 F1 cross: BbNn X BbNn P: BBNN X bbnn F1p%: 100% Black, long furred F2p%: 56.25% black; normal furred dominant; dominant 18.75% black; short furred dominant; recessive 18.75% brown; normal furred recessive; dominant 6.25% brown; short furred recessive; recessive F2g%: 6.25% BBNN 12.5% BBNn 12.5% BbNN 25% BbNn 6.25% BBnn 12.5% Bbnn 6.25% bbNN 12.5% bbNn 6.25% bbnn

  10. Page 6 #5 P: IAIBRr X IBirr F1g%:12.5% of each F1p%: 12.5% AB+ 12.5% AB- 25% B+ 25% B- 12.5% A+ 12.5% A-

  11. Page 6 #6 P: IAiRr X IBiRr F1p%: 18.75% AB+ 6.25% AB- 18.75% B+ 6.25% B- 18.75% A+ 6.25% A- 18.75% O+ 6.25% O-

  12. Page 7 #1 P: BBDD X bbdd 4 capitals = black 3 capitals = dark brown 2 capitals = light brown 1 capitals = dishwater blonde 0 capitals = blonde F1p%: 100% light brown F1 cross BbDd X BbDd F2p%: 6.25% black 25% dark brown 37.5% light brown 25% dishwater blonde 6.25% blonde

  13. Page 7 #2 4 capitals = black 3 capitals = dark brown 2 capitals = light brown 1 capitals = dishwater blonde 0 capitals = blonde F1p%: 12.5% dark brown 37.5% light brown 37.5% dishwater blonde 12.5% blonde

  14. Page 7 #3 6 capitals = 12 cm 5 capitals = 11 cm 4 capitals = 10 cm 3 capitals = 9 cm 2 capitals = 8 cm 1 capitals = 7 cm 0 capitals = 6 cm P FFGGLL X ffggll F1p%: 100% 9 cm F1 cross FfGgLl X FfGgLl F1p%: 1.563% 12cm 9.375% 11cm 23.44% 10cm 31.25% 9cm 23.44% 8cm 9.375% 7cm 1.563% 6cm

  15. Epistasis The C/c gene is epistatic to the B/b gene; fur can only have color if the C gene is present; so with a C a B gives black and with a C a b gives brown but with out a C the mouse is white

  16. p. 8 #1 B-black b-brown H-color h-no color P BbHh X BbHh F1p%: 56.25% black 18.75% brown 25% white

  17. p. 8 #2 P BbHh X bbHh B-black b-brown H-necessary for color h-no color F1p%: 37.5% black 37.5% brown 25% white #3 Both A and B are needed for colored flowers Aabb X aaBb F1p%: 25% colored flowers 75% white flowers

  18. In apples, green fruit is dependent on a dominant gene (G) and colored fruit , (yellow or red) on the recessive gene (g). In the presence of gg and a dominant gene R the color is red but when (R) is not present, that is (rr) the color is yellow. A green-fruited (GGRR) is crossed with a yellow-fruited (ggrr) tree. Give the F1 and F2 genotypic and phenotypic %’s. F1 cross GgRr X GgRr F1p%: 100% Green fruit F2p%: 75% green-fruited 18.75% red-fruited 6.25% yellow-fruited

  19. Environmental effect on phenotype the phenotypic range for a specific genotype is the norm of reaction Most human traits are multifactorial and are influenced by many factors both genetic and environmental

  20. Recessively Inherited Disorders Cystic fibrosis-defective chlorine channels in cell membranes causes thicken mucus survive into 20’s Tay-Sachs disease-dysfunctional enzyme that fails to break down a specific lipid in the brain; symptoms begin at a few months of age and usually death occurs by age of 2 Sickle-cell disease-hemoglobin molecules become fibrous with slight pressure or temperature changes; rbc’s sickle and block capillaries and cause excruciating pain; shows pleiotropy (many other problems); heterozygotes are immune to malaria Phenylketonuria-defective enzyme that breaks down the amino acid phenylalanine; thus it builds up in blood and causes mental retardation; children can be put on a special diet and will not be affected; laws enforce newborn screenings because it is the only treatable disorder

  21. pleiotropy

  22. Dominantly Inherited Disorders Huntington’s disease-Degenerative disease of the nervous system; symptoms do not show up until individual is 35-45 years old Achrondroplasia-a form of dwarfism; cartilage cells do not divide and bones do not grow

  23. p. 8 #5 P RRpp X rrPP RR or Rr = rose comb rrpp = single comb PP or Pp = pea comb R & P together = walnut comb F1p% 100% walnut comb F1 cross RrPp X RrPp F2p%: 56.25% walnut comb 18.75% rose comb 18.75% pea comb 6.25% single comb

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