Heredity and Genetics

1. Heredity and Genetics Mr. Erick Santizo

2. DEFINE the following • Allele • Chromosome • Metaphase • DNA • Interphase

3. Differentiate Mitosis Meiosis

4. Differentiate Mitosis Meiosis Either occurs in reproductive cells only or occurs in formation of gamate only # of chromosomes is halved in the daughter cells (haploid = n) Daughter cells are genetically different to parent cell and each other Four daughter cells are formed Exchange of genetic material between chromosomes • Occuris in body cells or somatic cells • # of chromosomes remains the same in the daughter cells (2n) diploid • Daughter cells are identical to parent cells and each other • Two daughter cells are formed • No exchange of genetical material between chromosomes

5. Variation • Term used to show differences between individuals of the same species. • Individuals of the same species varies • Differences seen in their physical appearance, such that no two are alike. • Even twins has subtle differences • Results from the “GENOTYPE” the genetic make up

6. Types of Variation • Continuous variation: anything that can be measured: eg. Human Height, foot length, skin color, leaf size and pod size in legumes. • Discontinuous variation: the differences are separated and clear cut; they do not merge or grade into each other. Examples are tongue-rolling in humans and the presence or absence of horns in cattle.

7. List as Continuous or discontinuos • A. Ear size • B. Brown Fur • C. Green Eyes • D. Wing span • E. Eyebrow Thickness • F. Widow’s peak • G. Elephant’s trunk

8. Genes • Pass on from parents to offspring • Different kinds of genes: for eye color, hair color, hair texture, shape of nose, size of lip, length of finger, length of arm and the list goes on. • There are 23 pairs of chromosomes: 23 that are from the father or paternal and 23 from the mother which is maternal. Pairs are called homologous chromosomes. (the same kind).

9. Chromosomes are made of genes/ units of inheritance. • * this is what controls specific characteristics in the organism. • Homologous pair: carries same set of genes; body cells has two copies of each gene HOWEVER • A gene can have more than one FORM called an allele.

10. ALLELES • So the cell may have two alleles for a gene that are the same as each other, or two alleles for a gene that are different. • If the alleles of a gene are the same: Homozygous • If the alleles of a gene are different: Heterozygous

11. Genotype Phenotype Black hair Black hair: B is dominant to b Red hair observable characteristics of the organism. Phenotype determined by genotype • BB : homozygous • Bb : heterozygous • bb : homozygous • Not visible; genetic make up

12. 2. For each of these genes, state which allele must be dominant and which must be recessive. Tt Dominant: ___ Recessive: ___ Gg Dominant: ___ Recessive: ___ Pp Dominant: ___ Recessive: ___ VV Dominant: ___ Recessive: ___ jj Dominant: ___ Recessive: ___ Answer the following

13. State whether each of these genotypes is homozygous dominant, heterozygous, or homozygous recessive. • HH _______________________________ • kk _______________________________ • Uu _______________________________ • vv _______________________________ • Bb _______________________________ • RR _______________________________

14. 4. For each of these genes, give the genotype that would be homozygous dominant, heterozygous, and homozygous recessive. • Y Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___ • Q Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___ • E Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___ • M Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___ • F Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___

15. 5. In each of these scenarios, give information about what the individual’s phenotype will be. • a. With respect to the gene for coat color in mink, if the allele U produces a yellow coat and the allele u produces a brown coat, then an individual that is UU will have this coat color: _______. • b. With respect to the gene for wing shape in house flies, if the allele I produces rounded wings and the allele i produces crooked wings, then an individual that is ii will have this wing shape: ________. • c. With respect to the gene for eye color in Pacific salmon, if the allele P produces light eyes and the allele p produces dark eyes, then a salmon with the genotype Pp will have this phenotype: ___________.

16. d. With respect to the gene for flipper length in bottlenose dolphins, if the allele T produces stunted non-functional flippers and the allele t produces normal flippers, then a dolphin with the genotype tt will have this phenotype: ________________. • e. Huntington’s Disease in humans is caused by the manufacture of a damaged version of a protein called huntingtin. We will use M for the damaged version of the protein huntingtin and the allele m produces normal huntingtin. If a person has the genotype Mm, will they have Huntington’s Disease? ________

17. 6. Given information about these phenotypes, determine which alleles are which. • a. In a group of mice, all individuals that are YY have long whiskers and all individuals that are yy have short whiskers. What is the allele for long whiskers? ____ What is the allele for short whiskers? ___ • b. In a group of fish, all individuals that are OO have thin fins and all individuals that are oo have thick fins. What is the allele for thin fins? ____ What is the allele for thick fins? ___ • c. In a group of dogs, all individuals that have the genotype Ee have floppy ears. What is the allele for floppy ears? ___ What is the allele for non-floppy ears? ___ • d. We will use the alleles G and g for the gene for coat spots in rabbits. All rabbits that have one of these alleles have spots. No rabbits without spots have that allele, they only have the other allele. What is the allele for spots? ____ What is the allele for no spots? ____ • e. We will use the alleles Q and q for the gene for fur smoothness in squirrels. All squirrels that are heterozygous have smooth fur, whereas only squirrels that are homozygous for one of those alleles have fluffy fur. Which is the allele for smooth fur? ___ Which is the allele for fluffy fur? ___

18. 7. In humans, who is heterozygous with respect to their biological sex? Men or women? Explain. ____________________________________________________ • 8. Suppose that 98% of the ferret population has smooth fur, and only 2% of the population has fluffy fur. Does this mean that the allele for smooth fur must be dominant? Explain. ___________________________________________________

19. Genetic Diagrams • -Shows the cross of two Genotypes. It shows the genotypes and phenotypes of the parents and the possible genotypes and phenotypes of the offspring. • Example: a genetic cross between homozygous black hair and a red hair parent: BB x bb • A.)All offspring would be heterozygous with black hair: Bb • B.) Heterozygous cross with homozygous gives 1Bb:1bb ratio 50%-50%

20. When two heterozygous cross together the ratio is- • 1red hair: 3 black phenotypic ratio; 25% of offspring having red hair • Genotypic ratio-1BB: 2Bb:1bb

21. Question #1 • Albinism (absence of pigmentation) in man is caused by a recessive gene which is transmitted in a normal fashion. A phenotypically normal (non-albino) couple have four children, the first three are normal and the fourth is albino. • i. What can you say about the genotype of the parents • ii.What is the possibility that their next child will be albino • iii. One of the normal children marries a normal woman. What predictions can be made of their First child? • Iv. The albino child marries a normal woman. What predictions can be made of their first child?

23. Question #2 • A breed of dogs has long hair dominant over short hair. A long-haired BITCH was first mated with a short-haired dog and produced three long-haired and three short haired puppies. Her second mating, with a long haired dog, produced a litter with all the puppies long haired. Use the symbol L to represent the allele for long hair and “l” to represent for short hair. • I. What was the genotype of the long-haired bitch? • ii. How could it be determined which of the long haired puppies of the second mating were homozygous?

25. Incomplete dominance • Shows blending; combination of expression of both alleles in the heterozygous condition. ( production of a new allele. Eg. Blue and yellow=green) • Eg. Red (RR) and White (rr) Flowers form pink flowers (Rr)

26. When crossing a Two pink flowers ( Rw) • Phenotypic ratio: 1red:2pink:1white • Genotypic ratio: 1RR: 1RW: 1WW • 2. In some cats the gene for tail length shows incomplete dominance. Cats with long tails and cats with no tails are homozygous for their respective alleles. Cats with one long tail allele and one no tail allele have short tails. For each of the following construct a punnett square and give phenotypic and genotype ratios of the offspring. 
a) a long tail cat and a cat with no tail 
b) a long tail cat and a short tail cat 
c) a short tail cat and a cat with no tail 
d) two short tail cats.

27. 2. long tails (L), no tails (ll), short tails (Ll). 
a) a long tail cat and a cat with no tail 
LL X ll, gametes for LL (L), gametes for ll (l), resulting phenotype ratio: 100% short tails, resulting genotype ratio: 100% Llb) a long tail cat and a short tail cat 
LL X Ll, gametes for LL (L), gametes for Ll (L & l), resulting phenotype ratio: 50% long tails / 50% short tails, resulting genotype ratio: 50% LL / 50% Llc) a short tail cat and a cat with no tail 
Ll X ll, gametes for Ll (L & l), gametes for ll (l), resulting phenotype ratio: 50% short tails / 50% no tails, resulting genotype ratio: 50% Ll / 50% lld) two short tail cats 
Ll X Ll, gametes for Ll (L&  l), gametes for Ll (L & l), resulting phenotype ratio: 25% Long Tailed, 50% short tailed, and 25% no tailed, resulting genotype ratio: 25% LL / 50% Ll / 25% ll

28. Co-dominance • A condition in which the alleles of a gene pair in a heterozygote are fully expressed thereby resulting in offspring with a phenotype that is neither dominant nor recessive.
SupplementA typical example showing codominance is the ABO blood group system. For instance, a person having A allele and B allele will have a blood type AB because both the A and B alleles are codominant with each other.

29. In some cattle the genes for brown hair (B) and for white hair (W) are co-dominant. Cattle with alleles for both brown and white hair, have both brown and white hairs. This condition gives the cattle a reddish color, and is referred to as Roan (BW). For each of the following construct a punnett square and give phenotypic and genotype ratios of the offspring. • a) a roan cow and a white bull • b) a brown cow and a roan bull • c) a white cow and a roan bull • d) a roan cow and a roan bull

30. brown hair (B), white hair (W), roan (BW). • a) a roan cow and a white bull BW X WW, gametes for BW (B & W), gametes for WW (W), resulting phenotype ratio: 50% roan / 50% white, resulting genotype ratio: 50% BW / 50% WW • b) a brown cow and a roan bull BB X BW, gametes for BB (B), gametes for BW (B & W), resulting phenotype ratio: 50% brown / 50% roan, resulting genotype ratio: 50% BB / 50% BW • c) a white cow and a roan bull WW X BW, gametes for WW (W), gametes for BW (B & W), resulting phenotype ratio: 50% roan / 50% white, resulting genotype ratio: 50% WW / 50% BW • d) a roan cow and a roan bull BW X BW, gametes for BW (B & W), gametes for BW (B & W), resulting phenotype ratio: 25% brown / 50% roan / 25% white, resulting genotype ratio: 25% BB / 50% BW / 25% WW

31. Blood groups AA 1.What are possible blood groups of children Whose parents are blood group “A” Heterozygous and “B” homozygous? AO BB BO AB b. What if both had heterozygous genotypes? OO

32. Sex determination Phenotype of parents: Male x Female Genotype: XY x XX Gametes: Offspring genotype: XX XX XY XY Offspring phenotype: female female male male Ratio: 1 female: 1 Male 50%-50% X Y X x

33. Sex-linked characteristics • The sex chromosomes carry genes other than those which determine sex. “ THE CHARACTERISTICS ARE SAID TO BE SEX-LINKED” Example: Haemophilia and colourblindness. Haemophilia or bleeder’s disease: the dominant allele “H” causes blood to clot NORMALLY. But the recessive allele “h” Causes HAEMOPHILIA.

34. Haemophilia H H H h h h H h

35. Question#1 1. A normal man married a normal woman and all the female offspring were normal, but about half of the male offspring were colour-blind and the other half were normal. How do you account for this (use knowledge of haemophilia)

36. Sickle cell anemia • The red blood cell can take a sickle shape instead of the normal biconcave shape. NN SS NS

37. Question • Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to see if they have the sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.

38. Pedigree charts • shows occurrence of a particular characteristic in a family tree. The chart can be used to show the possible genotypes of individuals in the chart, which can be important in genetic counseling.