Genetics Review Powerpoint MENDELS PEA EXPERIMENTS P 1

Genetics Review Powerpoint

MENDEL’S PEA EXPERIMENTS P 1 = Parental F 1 = Filial (pffspring) F 2 = Filial (pffspring) http: //hus. yksd. com/distanceedcourses/YKSDbiology/lessons/Fourth. Quarter/Chapter 11/11 -1/images/Mendel. Experiment. gif

Refresh your “Bio Brain” about: GENETICS VOCAB DOMINANT: RECESSIVE: HOMOZYGOUS (pure-breeding): HETEROZYGOUS (hybrid): GENOTYPE: PHENOTYPE:

Refresh your “Bio Brain” about: GENETICS VOCAB DOMINANT: gene that hides another represented by capital letter RECESSIVE: gene that is hidden by another represented by lower case letter

Refresh your “Bio Brain” about: GENETICS VOCAB HOMOZYGOUS (pure-breeding): Organism with two identical alleles for a gene TT OR tt HETEROZYGOUS (hybrid): Organism with two different alleles for a gene Tt

Refresh your “Bio Brain” about: GENETICS VOCAB GENOTYPE: Genetic make up of an organism “What genes it has” PHENOTYPE: Appearance of an organism “Way it looks”

Probability is the likelihood that an event will occur It can be written as a: Fraction ____ 1/4 25% Percent ____ Ratio ____ 1: 3

http: //www. arborsci. com/Cool. Stuff/Coin. Flip. jpg COIN FLIP There are 2 possible outcomes: HEADS TAILS The chance the coin will land on either one is: 1/2 ____ 1: 1 NOT 1: 2! 50% ____ Alleles segregate randomly just like a coin flip. . . So can use probability to predict outcomes of genetic crosses.

IN PEAS: R = round r = wrinkled T = tall t = short Y = yellow peas y = green peas P = purple flowers p = white flowers

Make a cross between a HOMOZYGOUS ROUND pea plant X PURE-BREEDING WRINKLED pea plant Show probabilities for: genotypes and phenotypes of possible offspring R R r Rr Rr Rr GENOTYPE ______ Rr PHENOTYPE ______ ROUND _______ 100% 4/4 show dominant phenotype _______ 0% 0/4 show recessive phenotype

Make a cross between a HOMOZYGOUS TALL pea plant X HOMOZGYOUS SHORT pea plant Show probabilities for: genotypes and phenotypes of possible offspring T T t Tt Tt Tt GENOTYPE ______ Tt PHENOTYPE ______ TALL _______ 100% 4/4 show dominant phenotype _______ 0% 0/4 show recessive phenotype

WHAT’s THE PATTERN? HOMOZYGOUS DOMINANT pea plant X HOMOZGYOUS RECESSIVE pea plant DO ONE IN YOUR HEAD HOMOZYGOUS PURPLE FLOWER X HOMOZGYOUS WHITE FLOWER pea plants ALL Pp 100% PURPLE FLOWERS 0% WHITE FLOWERS

IN PEAS: R = round r = wrinkled T = tall t = short Y = yellow peas y = green peas P = purple flowers p = white flowers

Make a cross between a HETEROZYGOUS YELLOW pea plant X HYBRID GREEN pea plant Show probabilities for: genotypes and phenotypes of possible offspring Y y Y YY Yy Yy y yy GENOTYPE PHENOTYPE YELLOW ______ YY Yy ____________ YELLOW ____________ yy GREEN 75% 3/4 show dominant phenotype _______ 25% 1/4 show recessive phenotype 1: 2: 1 ______ genotypic ratio 3: 1 _____ phenotypic ratio

Make a cross between a HETEROZYGOUS TALL pea plant X HETEROZYGOUS SHORT pea plant Show probabilities for: genotypes and phenotypes of possible offspring T t T TT Tt Tt t tt 1: 2: 1 ______ genotypic ratio GENOTYPE PHENOTYPE TALL ______ TT ____________ Tt TALL ____________ tt SHORT 75% 3/4 show dominant phenotype _______ 25% 1/4 show recessive phenotype _____ 3: 1 phenotypic ratio

WHAT’s THE PATTERN? HOMOZYGOUS DOMINANT pea plant X HOMOZGYOUS RECESSIVE pea plant DO ONE IN YOUR HEAD HOMOZYGOUS PURPLE FLOWER X HOMOZGYOUS WHITE FLOWER pea plants ______ 75% PURPLE? ______ 25% WHITE? _____ 25% PP? _____ 50% Pp? _____ 25% pp? 1: 2: 1 genotypic ratio _____ 3: 1 phenotypic ratio _______

POSSIBLE PARENT GAMETES? _____________ INDEPENDENT ASSORTMENT

POSSIBLE PARENT GAMETES? TTRR ____________ TR TR INDEPENDENT ASSORTMENT

POSSIBLE PARENT GAMETES? ttrr ____________ tr tr INDEPENDENT ASSORTMENT

TR TR TR tr Tr. Rr tr Tt. Rr TR Tr. Rr Tt. Rr GENOTYPE Tr. Rr _______ tr Tt. Rr PHENOTYPE TALL & ROUND ______ Tt. Rr tr Tt. Rr

POSSIBLE PARENT GAMETES? Tt. Rr ____________ TR tr t. R Tr INDEPENDENT ASSORTMENT

POSSIBLE PARENT GAMETES? Rr. Yy ____________ RY ry r. Y Ry INDEPENDENT ASSORTMENT

RY RY Ry r. Y ry ____ Round & Yellow Ry ____ Round & green r. Y ____ Wrinkled & yellow ry ____ wrinkled & green

RY Ry r. Y ry RRYY RRYy Rr. YY Rr. Yy ____ Round & Yellow Ry RRYy RRyy Rr. Yy ____ Round & green r. Y Rr. Yy rr. YY rr. Yy ____ Wrinkled & yellow ry Rr. Yy Rryy rr. Yy ____ wrinkled & green RY Rryy rryy

RY Ry r. Y ry RRYY RRYy Rr. YY Rr. Yy ____ 9 Round & Yellow Ry RRYy RRyy Rr. Yy Rryy 3 Round & ____ green r. Y Rr. Yy rr. YY rr. Yy 3 Wrinkled ____ & yellow rryy 1 wrinkled ____ & green RY ry Rr. Yy Rryy rr. Yy Sign of HETEROZYGOUS DIHYBRID cross What’s the pattern?

____ 9 ______ dominant TRAIT 1 ; ______ dominant TRAIT 2 3 ______ dominant TRAIT 1; _______ recessive ____ TRAIT 2 recessive TRAIT 1; _______ dominant 3 ______ TRAIT 2 1 ______ recessive TRAIT 1; _______ recessive 9: 3: 3: 1 _____ratio is a clue that it’s a ______________cross HETEROZYGOUS TWO gene

NON-MENDELIAN INHERITANCE COMPLETE DOMINANCE http: //www. emc. maricopa. edu/faculty/farabee/BIOBK/Bio. Book. TOC. html INCOMPLETE DOMINANCE F 2 generation- Not a 3: 1 ratio Heterozygote= blended intermediate phenotype

NON-MENDELIAN INHERITANCE CO-DOMINANCE HETEROZYGOTE: Both traits are expressed together (NO BLENDING) Roan horse has BOTH red & white hair (NOT A PINK HORSE! An A allele AND a B allele make BOTH A and B GLYCOPROTEINS = AB Blood type

Make a cross between TWO HETEROZYGOUS ROAN HORSES Show probabilities for: genotypes and phenotypes of possible offspring R W R RR RW GENOTYPE PHENOTYPE ______ RR 25% 1/4 RED ____________ RW 50% 1/2 ROAN ____________ WW 25% 1/4 WHITE W RW WW 1: 2: 1 ______ genotypic ratio _____ 1: 2: 1 phenotypic ratio

Make a cross between a HOMOZYGOUS RED HORSE and HOMOZYGOUS WHITE HORSE Show probabilities for: genotypes and phenotypes of possible offspring R R W RW RW GENOTYPE ______ RW PHENOTYPE ______ ROAN 100% ROAN offspring _____

NON-MENDELIAN INHERITANCE BLOOD TYPE MULTIPLE ALLELE TRAIT & CODOMINANT TRAIT ALLELES: _____ allele is dominant to _____ allele. _____ and _____ are CODOMINANT. BOTH SHOW TOGETHER

NON-MENDELIAN INHERITANCE BLOOD TYPE MULTIPLE ALLELE TRAIT & CODOMINANT TRAIT ALLELES: ____ A B ____ O ____ A allele is dominant to _____ O allele. B allele is dominant to _____ A and _____ B are CODOMINANT. BOTH SHOW TOGETHER _____

SEX-LINKED GENES Some genes are carried on SEX chromosomes X-linked recessive disorders: Hemophilia Color blindness Duchenne Muscular Dystrophy FEMALES XX= ________ MALES XY = ________ X-linked traits show up more frequently in males No backup X to cover for the “broken” gene

X-LINKED Cross a colorblind male with a normal vision (non-carrier) female XC XC c X Y C c X X C X Y X CX c X CY PHENOTYPE 100% normal vision 50% normal males 50% carrier females

X-LINKED Cross a hemophilia male with a carrier female XH Xh h X Y H h X X H X Y X h. Y PHENOTYPE 25%- hemophilia female 25%- hemophilia male 25%- normal male 25% - normal (carrier) females

TEST CROSSES

TEST CROSSES Dominant looking parent could have these genotypes: TT Tt ____ OR _____ Can’t tell which by looking. Test cross used to determine which it is. ALWAYS TESTCROSS WITH A HOMOZYGOUS RECESSIVE _________________ Offspring provide clue about genotype of unknown parent.

TEST CROSSES T T t Tt Tt All offspring will be TALL T t t Tt tt tt 50% will be TALL 50% will be SHORT If any offspring show the recessive trait. . . unknown parent genotype was ____. If all offspring show the dominant trait. . . still don’t know. BOTH genotypes could produce offspring that look dominant! Tt

Deafness in dogs is caused by a recessive allele. Deaf dogs have the genotype dd. You have a hearing dog. Do a test cross to determine its genotype. D D d Dd Dd D d d Dd dd An actual test cross results in a litter with: 5 hearing puppies 0 deaf puppies What can you conclude?

Firebreathing (F) in dragons is dominant over NON-firebreathing (f). You have a fire-breathing dragon. What possible alleles could the fire-breathing parent have? _______ OR ____ EXPLAIN how you could use a TEST CROSS to help determine the parental genotype. Show the results of test crossing BOTH OF THE POSSIBLE PARENT GENOTYPES: An actual test cross results in a litter with: 6 firebreathing dragons 1 NON-firebreather EXPLAIN how you could use these results to determine the correct parental genotype.

TEST CROSS Used to determine genotype of unknown DOMINANT LOOKING parent Always cross with HOMOZYGOUS RECESSIVE (EX: tt) Observe offspring. If any offspring show recessive trait… know parent was HETEROZYGOUS If all offspring show DOMINANT trait. . . Still don’t know genotype. Do another test cross. Both TT and Tt can produce tall offspring with tt cross

DRAGONS An actual test cross results in a litter with: 6 firebreathing dragons 1 NON-firebreather F F f Ff Ff Ff F f f Ff ff What can you conclude? Unknown parent genotype is Ff Only way you can get a NON-FIREBREATHER

Write a NULL hypothesis that describes the mode of inheritance for the trait (purple eyes) THERE IS NO DIFFERENCE BETWEEN THE OBSERVED DATA AND THE EXPECTED DATA IF PURPLE EYES IS A(N) _________ TRAIT I would expect this pattern in the F 1 offspring ________________ I would expect this pattern in the F 2 offspring ________________

p + + + p +p +p F 1 100% wild type + p p ++ +p +p pp F 2 75% wild type 25% purple eyed IF YOU IGNORE SEX: If 1000 flies, expect 750 to be wild type and 250 to be purple eyed If you DON’T IGNORE SEX: 375 WT MALES 375 WY FEMALES 125 purple eyed FEMALES Calculate Chi-square DO YOU ACCEPT OR REJECT THE NULL HYPOTHESIS?

H 0 - There is no difference between the frequencies observed and the frequencies expected if PURPLE eyes is an autosomal recessive trait.

m m + +m +m + + ++ IF GENE is AUTOSOMAL and RECESSIVE TO + MALES: FEMALES 1: 1 F 1 All = +m (wildtype) m +m mm F 2 ¼ = ++ ½ = +m ¼ = mm 75% wildtype 25% mutant 3: 1 http: //www. exploratorium. edu/exhibits/mutant_flies/curly-wings. gif

M M + +M +M + + M ++ IF GENE is AUTOSOMAL and DOMINANT TO + MALES: FEMALES 1: 1 F 1 All = +M (mutant) M +M +M MM F 2 ¼ = ++ ½ = +M ¼ = MM 25% wildtype 75% mutant 1: 3

IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad Mutant mom X wild type dad F 1 + Xm y X m X + X my Xm X + X my 50% normal females 50% mutant males X+ X+Xm X+y Xm X my 25% normal females 25% mutant males 25% normal males When using Virtual fly lab Choose ignore sex and see if it changes the ratios

POSSIBLE PARENT GAMETES? Rr. Yy ____________ RY ry r. Y Ry INDEPENDENT ASSORTMENT

RY Ry r. Y ry RRYY RRYy Rr. YY Rr. Yy ____ 9 Round & Yellow Ry RRYy RRyy Rr. Yy Rryy 3 Round & ____ green r. Y Rr. Yy rr. YY rr. Yy 3 Wrinkled ____ & yellow rryy 1 wrinkled ____ & green RY ry Rr. Yy Rryy rr. Yy Sign of HETEROZYGOUS DIHYBRID cross What’s the pattern?

If CLOSE TOGETHER on homologous chromosomes, STAY TOGETHER and end up together 100% of time Act like one gene

http: //image. slidesharecdn. com/meiosisnotes-100204185918 -phpapp 02/95/meiosis-notes-23 -728. jpg? cb=1265310007 CROSSING OVER If far apart on homologous chromosomes, end up together 50% of time RECOMBINANTS- Put different maternal/paternal alleles together on different chromosomes

http: //image. slidesharecdn. com/meiosisnotes-100204185918 -phpapp 02/95/meiosis-notes-23 -728. jpg? cb=1265310007 INDEPENDENT ASSORTMENT If on different chromosomes, END UP TOGETHER 50% of time.

____ 9 ______ dominant TRAIT 1 ; ______ dominant TRAIT 2 3 ______ dominant TRAIT 1; _______ recessive ____ TRAIT 2 recessive TRAIT 1; _______ dominant 3 ______ TRAIT 2 1 ______ recessive TRAIT 1; _______ recessive 9: 3: 3: 1 _____ratio is a clue that it’s a ______________cross HETEROZYGOUS TWO gene

A large ear of corn has a total of 433 kernals, including 271 purple & starchy, 73 purple & sweet, 63 yellow & starchy, and 26 yellow & sweet. HYPOTHESIS: This ear of corn was produced by a dihybrid cross (Pp. Ss X Pp. Ss) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9: 3: 3: 1. Test your hypothesis using Chi-square and probability values. SHOW YOUR WORK!

H 0 - There is no difference between the frequencies observed and the frequencies expected for a HETEROZYGOUS DIHYBRID (9: 3: 3: 1) cross. TOTAL = 433 offspring IF 9: 3: 3: 1 then expect: Purple & Starchy = 433 X 9/16 = 243. 56 Purple & Sweet = 433 X 3/16 = 81. 19 Yellow & Starchy = 433 X 3/16 = 81. 19 Yellow & Sweet = 433 X 1/16 = 27. 06 271 73 63 26 243. 56 81. 19 27. 06 4 -1=3 27. 44 -8. 19 -1. 06 752. 95 67. 08 330. 88 1. 12 3. 09 0. 83 4. 08 0. 04 8. 04 is larger than 7. 82 REJECT THE NULL There is a difference between observed and expected for 9: 3: 3: 1 = NOT A 9: 3: 3: 1 cross

What is the probability? Aa. Bb. Cc. Dd parent genome What is the probability of producing a gamete with this gene combination? ABCD _______________ a. Bc. D ________________

What is the probability? Aa. Bb. Cc. Dd parent genome What is the probability of producing a gamete with this gene combination? ½ X ½ X ½ = 1/16 ABCD _______________ ½ X ½ X ½ = 1/16 a. Bc. D ________________

What is the probability? Aa. Bb. Cc. DD parent genome What is the probability of producing a gamete with this gene combination? abc. D _______________ abcd ________________ Abc. D _______________

What is the probability? Aa. Bb. Cc. DD parent genome What is the probability of producing a gamete with this gene combination? ½ X ½ X 1 = 1/8 abc. D _______________ ½X½ X½X 0 =0 abcd _______________ ½ X ½ X 1 = 1/8 Abc. D _______________

What is the probability? Aa. Bb. Cc. Dd X Aa. Bb. Cc. Dd parents What is the probability of producing a offspring with this gene combination? aabbcc. Dd _______________ Aa. BBcc. DD_______________ Aa. BBCCDd________________

What is the probability? Aa. Bb. Cc. Dd X Aa. Bb. Cc. Dd parents What is the probability of producing a offspring with this gene combination? aabbccdd _______________ ¼ X ¼ X ¼ = 1/256 ½ X¼ X¼ X¼ = 1/128 Aa. BBcc. DD_______________ ½ X ¼ X ½ = 1/64 Aa. BBCCDd________________

What is the probability? Aa. Bb. Cc. Dd X Aa. Bb. Cc. Dd parents What is the probability of producing a offspring with this gene combination? AABb. Cc. Dd _______________ Aa. BBccdd_______________ Aa. Bb. Cc. Dd________________

What is the probability? Aa. Bb. Cc. Dd X Aa. Bb. Cc. Dd parents What is the probability of producing a offspring with this gene combination? AABb. Cc. Dd _______________ ¼ X ½ X ½ = 1/32 ½ X ¼ X ¼ = 1/128 Aa. BBccdd_______________ ½ X ½ X ½ =1/16 Aa. Bb. Cc. Dd________________

PEDIGREES = male; doesn’t show trait = female; doesn’t show trait = shows trait = carrier; doesn’t show trait

1 3 2 4 Circle all males that show the trait BLUE Circle all females that show the trait RED Circle all carrier females GREEN Circle all carrier males PURPLE If this is an autosomal recessive trait, what is the genotype of individual #1? If this is an autosomal recessive trait, what is the genotype of individual # 2? If this is an autosomal recessive trait, what is the genotype of individual #3?

1 3 2 4 Circle all males that show the trait BLUE Circle all females that show the trait RED Circle all carrier females GREEN Circle all carrier males PURPLE If this is an autosomal recessive trait, what is the genotype of individual #1? If this is an autosomal recessive trait, what is the genotype of individual # 2? If this is an autosomal recessive trait, what is the genotype of individual #3?

Write the genotype of each individual next to the symbol. Is it possible that this pedigree is for an autosomal recessive trait? Is it possible that this pedigree is for an X-linked recessive trait?

Write the genotype of each individual next to the symbol. Aa Aa Aa or AA aa Is it possible that this pedigree is for an autosomal recessive trait? YES Is it possible that this pedigree is for an X-linked recessive trait?

Write the genotype of each individual next to the symbol. XB Y XB X b XB Y Xb X b Is it possible that this pedigree is for an autosomal recessive trait? Is it possible that this pedigree is for an X-linked recessive trait? NO

The inheritance of the disorder in II-3 from his father rules out what form of inheritance? Xb Y Can’t be X-linked recessive Males get their X-linked allele from their mother If dad passed to son it must be AUTOSOMAL http: //www. mybookezzz. org/ebook. php? u=a. HR 0 c. Dov. L 21 j. Yi 5 i. ZXJr. ZWxle. S 5 l. ZHUv. Y 291 cn. Nlcy 9 t. Y 2 I 0 MS 9 wcm. Fjd. Glj. ZV 9 wcm 9 ib. GVtc 19 hbn. N 3 Ln. Bk. Zgp. Qcm. Fjd. Glj. ZSBwc m 9 ib. GVtcy. Aod 2 l 0 a. CBhbn. N 3 ZXJz. KSBUa. Glz. IHRo. ZSBk. ZWdy. ZWUgb 2 Yg. ZGlm. Zmljd. Wx 0 e. SBv. Zi. Au. Li 4=

Is this trait inherited as AUTOSOMAL RECESSIVE? http: //www. mansfield. ohio-state. edu/~sabedon/biol 1128. htm#A 1

Is this trait inherited as AUTOSOMAL RECESSIVE? aa Aa aa aa AUTOSOMAL RECESSIVE POSSIBLE Aa aa aa aa Aa aa aa http: //www. mansfield. ohio-state. edu/~sabedon/biol 1128. htm#A 1 Aa aa

Is this trait inherited as AUTOSOMAL DOMINANT ? Aa aa Aa Aa AUTOSOMAL DOMINANT IMPOSSIBLE aa Aa Aa Aa aa Aa Aa http: //www. mansfield. ohio-state. edu/~sabedon/biol 1128. htm#A 1 aa Aa

Is this trait inherited as X-LINKED RECESSIVE? Xa Xa XA Y Xa Xa http: //www. mansfield. ohio-state. edu/~sabedon/biol 1128. htm#A 1 X-LINKED RECESSIVE IMPOSSIBLE

Is this trait inherited as X-LINKED DOMINANT? X-LINKED DOMINANT IMPOSSIBLE XA Xa Xa Y XA Y Xa Xa Xa Y XA X? http: //www. mansfield. ohio-state. edu/~sabedon/biol 1128. htm#A 1

PATTERNS ARE THE KEY Image modified from: http: //www. laskerfoundation. org/rprimers/gnn/timeline/1866. html http: //www. accessexcellence. org/AB/GG/mendel. html

MAKING PUNNETT PREDICTIONS http: //www. exploratorium. edu/exhibits/mutant_flies. html

USING CHI-SQUARE TO ANALYZE GENETICS DATA CROSS BETWEEN A RED EYED (WILD TYPE) FEMALE AND PURPLE EYED MALE DATA: F 1 - 570 WT females; 615 WT males F 2 - 460 WT females; 147 purple eyed females; 451 WT males; 138 Purple eyed males IGNORE SEX : 911 WT flies; 285 purple eyed flies pur + + p +p + + +p +p p http: //www. exploratorium. edu/exhibits/mutant_flies. html + p + + + p +p p p

USING CHI-SQUARE TO ANALYZE GENETICS DATA CROSS BETWEEN A RED EYED (WILD TYPE) FEMALE AND SEPIA EYED MALE DATA: F 1 - 529 WT females; 497 WT males F 2 - XXXX WT females; XXXX sepia eyed females; XXXXX WT males; XXXXX sepia eyed males + se se + se + + se se se http: //www. exploratorium. edu/exhibits/mutant_flies. html se +

H 0 - There is no difference between the frequencies observed and the frequencies expected if sepia eyes is an autosomal recessive trait.

Male yellow body Female WT P 1 t+ t+ Xt+ Y F 1 622 male WT 590 female WT F 2 608 female WT 326 male WT 281 male Yellow body http: //www. exploratorium. edu/exhibits/mutant_flies. html

y + + +y +y IF GENE is AUTOSOMAL and RECESSIVE TO + MALES: FEMALES 1: 1 +y +y F 1 All = +y (wildtype) + + y y ++ +y yy F 2 ¼ = ++ ½ = +y ¼ = yy 75% wildtype 25% yellow body 3: 1 http: //www. exploratorium. edu/exhibits/mutant_flies/curly-wings. gif

Y Y + +Y +Y + Y ++ +Y + Y +Y YY IF GENE is AUTOSOMAL and DOMINANT TO + MALES: FEMALES 1: 1 F 1 All = +Y (yellow body) F 2 ¼ = ++ ½ = +Y ¼ = YY 25% wildtype 75% mutant 1: 3

IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad Mutant mom X wild type dad F 1 + Xm y X m X + X my Xm X + X my 50% normal females 50% mutant males X+ X+Xm X+y Xm X my 25% normal females 25% mutant males 25% normal males When using Virtual fly lab Choose ignore sex and see if it changes the ratios

IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad Wild type mom X yellow body dad F 1 Xt y Xt+Xt+ Xt+y Xt+Xt Xt+y 50% normal females 50% normal males F 2 Xt+ y Xt+Xt+ Xt+y Xt Xt+Xt X ty 50% Wild type FEMALE 25% tan MALES 25% Wild type males When using Virtual fly lab Choose ignore sex and see if it changes the ratios

IF GENE is X-linked dominant MUTANT MOM X WILD TYPE DAD F 1 X+ y X M X + X My XM X + X My 50% mutant females 50% mutant males F 2 XM y X+ X +X M X +y XM X MX M X My 50% mutant females 25% wild type males 25% mutant males When using Virtual fly lab Choose ignore sex and see if it changes the ratios

IF GENE is X-linked dominant WILD TYPE MOM X MUTANT DAD F 1 XM y X +X M X +y X+ X +X M X +y 50% mutant females 50% normal males X+ F 2 y X+ X +X + X +y XM X + X My 25% wild type females 25% mutant females 25% wild type males 25% mutant males When using Virtual fly lab Choose ignore sex and see if it changes the ratios

If CLOSE TOGETHER on homologous chromosomes, STAY TOGETHER and end up together 100% of time Act like one gene

http: //image. slidesharecdn. com/meiosisnotes-100204185918 -phpapp 02/95/meiosis-notes-23 -728. jpg? cb=1265310007 CROSSING OVER If far apart on homologous chromosomes, end up together 50% of time RECOMBINANTS- Put different maternal/paternal alleles together on different chromosomes

http: //image. slidesharecdn. com/meiosisnotes-100204185918 -phpapp 02/95/meiosis-notes-23 -728. jpg? cb=1265310007 INDEPENDENT ASSORTMENT If on different chromosomes, END UP TOGETHER 50% of time.

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778 - wild type 785 - black-vestigial 158 - black- normal wings 162 - gray body-vestigial wings Is it 9: 3: 3: 1? (2 genes on 2 different chromosomes) Is it 3 Wild type : 1 black vestigial? (linked on homologous chromosomes)

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778 - wild type 785 - black-vestigial 158 - black- normal wings 162 - gray body-vestigial wings What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. Recombinants = OFFSPRING: Total 778 - wild type 320 = 17% 785 - black-vestigial 1883 158 - black- normal wings 162 - gray body-vestigial wings

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778 - wild type 785 - black-vestigial 158 - black- normal wings 162 - gray body-vestigial wings What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. Recombinants = OFFSPRING: Total 778 - wild type 314 = 16. 7% 785 - black-vestigial 1877 158 - black- normal wings 162 - gray body-vestigial wings What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for gray body and red eyes) is mated with a black fly with purple eyes. OFFSPRING: 721 - gray body/red eyes 751 - black body/purple eyes 49 - gray body/purple eyes 45 - black body/red-eyes What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for gray body and red eyes) is mated with a black fly with purple eyes. Recombinants = OFFSPRING: Total 721 - gray body/red eyes 751 - black body/purple eyes 94 = 6 % 1566 49 - gray body/purple eyes 45 - black body/red-eyes What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for normal bristles and red eyes) is mated with a spineless bristle fly with sepia eyes. OFFSPRING: 648 - normal bristles/red eyes 681 - spineless bristles/sepia eyes 72 - normal bristles/sepia eyes 83 - spineless bristles/red-eyes What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for normal bristles and red eyes) is mated with a spineless bristle fly with sepia eyes. Recombinants = OFFSPRING: Total 648 - normal bristles/red eyes 681 - spineless bristles/sepia eyes 155 = 10. 4% 1484 72 - normal bristles/sepia eyes 83 - spineless bristles/red-eyes What is the recombination frequency between these genes?

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 20% A-D 10% B-C 15% B-D 5%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 20% A-D 10% B-C 15% B-D 5%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 10% A-D 30% B-C 24% B-D 16%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 10% A-D 30% B-C 24% B-D 16%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 10% A-D 30% B-C 24% B-D 16%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 10% A-D 30% B-C 24% B-D 16% CABD

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-B 8% A-C 28% A-D 25% B-C 20% B-D 33%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-B 8% A-C 28% A-D 25% B-C 20% B-D 33%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-B 8% A-C 28% A-D 25% B-C 20% B-D 33%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-B 8% A-C 28% A-D 25% B-C 20% B-D 33%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-B 8% A-C 28% A-D 25% B-C 20% B-D 33%