Pre-Lab 16 Name: NOI\.MENDELIAN GENETICS AND HUMAN HEREDITY 1. If two genes are linked they are on the 5a vn e chromosome. 2. A male who exhibits a sex-linked trait, inherited it from which parent? nv\ie / The genes for sex-linked traits are carried on the X tlr*t ov^ *- Would you expect this disease to be found equally in males and females? rrl o 3. Define the following terms: a. Pleiotropy: 4q^-<-.t =J: rn*--=-\r^c^^. \ \the--"\tq b. Epistasis: 1-< * < e tt=--f s <rc "<:s, *!t -,'*lt* O* c. Co-dominanse: at\ o.,t\e\<\ a..:i= clo-^.,^ "*lcl d. Multiple allelism: qe.,".< c Vt rts 't n v,,^.<r-,'-< al\ett q Sq-r,..^{. <rc,,.-e t\a 165
T ; l Lab Report 16 Name NON.MENDELIAI\ GENETICS AND IIUMAN HEREDITY A. Exceptions to Mendelian principles 1. Flower color in snapdragons is controlledpy incomplete dominance tance. Complete the following cross. Use the genetic symbols: ), White allele (Cw) Phenotype: Genotype: Red flowers.(. -( (u- ) (ck@ CK ClE Pink flowers t CKc*l r c(; cwl a dcs c(ok c*j aqrl cfoul Phenotypic ratio: [: I Genotypic ratio: cgck : ckcs ( {r t) 2. The ABO blood gloups are controlled by multiple alleles inheritance. Complete the followiitg "tots using the geneticsymbols: IA, IB and i Phenotype: Genotype: AB blood type t tarf ) r la; tfl 1A X (t lai 16; O blood fype (t) t ( i&) lhi r8 i Phenotypic ratio: Genotypic ratio: t: I f I't\ IA'' :' lai 167 j
3. A male with blood type A, whose mother was type O and father was type AB, marries a woman who is type AB. Phenotype: Genotype: A blood type ( lai ) < ta; i l X AB blood type r la\g I rti; rfr ra ta ls th \n \kr $ IN't loi What is the probability that they will have a type A child? Probability: z '/. (e'() ' Probability = o/. What is the probability that they will have a type O child? 4. 5. A woman who is type A blood is married to a man with type AB blood. She has a child that is type O. Isthispossible? Ncrf ur=rtfsj s!'tc ntlf,fp Uf\}t /N)"ry*q*"N Possible genotype of the parents: x (Nl n I tjko t'{a\ A child is born with sickle cell anemia, Work out the cross predicting the genotypes of the parents to show how seemingly healthy parents can have a child with the disorder. rdn r tnj", I NI vj t Ni ^ I N) V1 168
-T 6. In sickle cell anemia, a mutation in the h e r, qlo b, ^ to become deformed. gene causes the red blood cells 7. 8. There are two alleles for the basic eye color: the dominant allele for making the brown pigment melanin and the recessive allele for blue color (absence of melanin). Explain how an individual can have dark brown or light brown eyes. f,rl.d.."tr\.!.-- -t e\i/7-(r.,*^ "F t-,.^.-r. c,.tor LT or.-^ob<- Sene If you look around, you will find that humans can be very tall to very short. Give a genetic explanation for the gradation found in human height. (o r;.--tr "Lo..,. 'r.*^ 44 B. Autosomal linked genes l. In a dihybrid, if two genes are linked, how does its gamete frequency linked gene? \ r tr\,,1"., [.rr h^-* differ from the non- In Drosophila,the wing shape and body color are often inherited together as the parental phenotypes. a. The genes for the wings and body color are referred to as Lt sk rca' genes. b. what can you say about the location of the genes for these traits? -il-'e c. The process that breaks this connection between the linked genes to form the new recombinants is f c c-r>*..l,'^ * ^ (q---,-rs -..$""1 169 -
tf{ D-,**& V-..^ \,6) tlr.rf 3. If ajltr:luid*qlong-winged, grey fly is mated *i,ft *5lack body fly.ihe number and phenotype of the offspring is shown in the following table. -Determine the per cent crossover by dividing the total number of recombinant offspring (long-winged, black body and vestigial wing, grey body) by the total number of offspring and then multiply by 100. Phenotype Long-winged, grey body Long-winged, black body Vestigial wing, grey body Vestigial wing, black body Number of Offspring Phenotype Ratio (divide by lowest zscl so Total progeny * To determine the per cent crossover, the total number of recombinant offspring is divided by the total number of offspring and then multiplied by 100. *Percent Crossover (Number of Map Units) 53+ sro -573 4. Answer the following questions based on this data from test cross between the dihybrid red- ^ Y- V eyed, long winged Drosophilawith the homozygous purple-eyed curved wing flies:,{'g E @qi9 t-+--/ I l{'eo Red tongwlngs eyes, long wings I I ' I {*.u Red e.yes, eyes, curved wings I ton * r x rtfrl Purple eyeso long wings Purple eyes, curved wings Number of Offspring /( \ofiu i a. Calculate the o/o crossover that occurred between the two genes' b. What is the map distance between the two genes? *2 I * "r, t70
6. C. 1. Are the dihybrid gametes produced in equal amounts? Sex-linkage A male inherits his X chromosome from NAq Vrl t- 2' A female inherits her X chromosomes from rr'o'rn o ''-L d.. n A 3. Work out the cross for sex-linked trait hemophilia where the recessive allele prevents the body from making clotting factors. Use the genetic symbols: normal clotting (Xn), k-*-- hemophilia (Xh). Phenotype: Mother - Carrier X Father - Normal Genotype: (XHXh ) ( XHY I Gametetypes: (XK,rtX\) ( XH I XX Z XY 9{ XH u. It'' qa^^ellt. --> Aa';LJrr ----> J O\A -What - is the probability that they will have a daughler with hemophilia? Probability: What is the probability that they will have a son with hemophilia? Probability : g'/, - /n"b.l..t,+1 ii'<1!t-.,^'. c* &y f _ rsz_ t,'rl hu^*{lr\,'*-l 171 1
4. Complete the pedigree for sex-linked trait colorblindness by giving the genotype for each family member. Use the following genetic symbols: Normal (Xc); Colorblindness (X9 XffieX X" * N*rnr*tvixtun ren'r*te il t{*rrn*$vr*ts*x nral* X C V X* x* W.fl*rlt,rh*ir*dfsnral* (^ ry f- Vta V {ol*rlriind male n \ fr (x.l ()EV (N't )f() otll (x'v xl (x"x.) 172