Smorgasbord Review 2 Key

Smorgasbord Review #2 NAME: KEY 1. Flower color in snapdragons can be red, white, or pink. Pink snapdragons occur when the genotype is heterozygous. Two pink snapdragons are crossed. What are the phenotypes of their offspring and in what proportion? → P: I Red: 2 pink: I white RR 2. Coat color in cattle can be red, white, or roan (both red and white). If 2 roan cattle are crossed, what are the phenotypes of their offspring and in what proportion? Al incomplete dominance 3. A rabbit breeder wants to get as many white-haired, red-eyed rabbits as possible. However, these are both recessive traits and the 2 parent rabbits have gray hair and black eyes. How many white-haired, red-eyed rabbits will result if both parents are heterozygous for both traits? ene: color gene white Bed: h gray: H gene: eye redie black: E I white hair + red eye rabbit 4. The allele for coat solor in cats is carried on the X chromosome. A calico cat has an allele for orange and an aliele for black. If a callso cat (female) and a black cat are crossed, what are the phenotypes of the offspring 5. Color-blindness is a sex-linked recessive disorder. If Don is not colorblind and his wife, Betty, is not colorblind, would it be possible for them to have children who are colorblind? Show the cross to prove it. IA i BB 1818 II IA 6. Two people, one heterozygpus-for-type-A-blood-and one who is homozygous for type B blood have children. What is the phenotypic ratip? 2 Type AB 2 Type B 7. The gene for colorblindness is carried on the X chromosome and is recessive. A man, whose father was colorblind, has a colorblind daughter. a.) Is this man colorblind? Yes How do you know? has to pass on the recessive allele + only has one x. b.) Where did he get his gene for colorblindness? c.) Must the fathers of all colorblind girls s be colorblind? Les wh •because they hav to be able to pass on the colorblind recessive Why? 8. Hornless (H) in cattle is dominant over horned (h). A homozygous hornless bull is mated with a homozygous horned cow. What will be the genotype and phenotype of the first generation? + P: 4 homiess: O homed 9. In some chickens, the gene for feather color is controlled by codominance. The allele for black is B and the allele for white is W. The heterozygous phenotype is known as erminette (black and white spotted). If two erminette chickens were crossed, what is the probability that: a. They would have a black chick225% b. They would have a white chick225% BBB BW WIBW IWW 10. In mice, the ability to run normally is a dominant trait. Mice with this trait are called running mice (R). The recessive trait causes mice to run in circles only. Mice with this trait are called waltzing mice (r). Hair color is also inherited in mice. Black hair (B) is dominant over brown hair (b). Cross a heterozygous running, heterozygous black mouse with a homozygous running, heterozygous black mousi je Bb x RRBb gene: speed ↓ G:2 RREB: 4RRBB: 2 RRbb: 2 RFBB: 4 Br Bb: 28rbb 01. & homin hoir R run: gene hair Black: B Brown: b Problem 11: Marfan Syndrome Inheritance: Autosomal dominant Genotype of Individual II-2: Hh (Heterozygous) Probability of child with Marfan Syndrome: 50% (Punnett square: Hh x hh) Problem 12: Familial Hypercholesterolemia Pedigree: A pedigree is drawn showing Jason, Michelle, and their parents. Genotypes: Jason: hh (homozygous recessive) Michelle: Hh (heterozygous) Michelle's Parents: Hh (both heterozygous) Probability of affected child: 0% The worksheet focuses on understanding genetic inheritance patterns, interpreting pedigrees, and using Punnett squares to predict probabilities of offspring genotypes and phenotypes.