1 The Big Picture: an outline of the concepts covered to date 1.Genes are physical units of hereditary that carry information from one generation to the next 2. Mendel elucidated the following principles regarding the inheritance patterns of genes  A. Each individual contains two copies of a given gene  B. Genes have different forms called alleles. The form that is expressed phenotypically in the heterozygote is known as the dominant allele. It is an operational definition  C. These copies (alleles) segregate from one another to form gametes (There is a single copy of each gene in a gamete)  D. Pairs of genes assort independently from one another during gamete formation 3. The inheritance pattern of genes parallels the behavior of chromosomes at meiosis. This generated the hypothesis that genes reside on chromosomes OR A a A a B b b B

2 The Big Picture 4. Exceptional patterns of chromosome segregation  Non-disjunction: homologous chromosomes migrate to the same pole during meiosis  The X/X and X/Y sex chromosomal system produces exceptional segregation patterns because males contain only one copy of X-linked genes 5. Exceptional expression patterns: Incomplete dominance, Co-dominance, Lethal alleles 6. Genes that reside close to one another on the same chromosome do not assort independently- linkage 7. Occasionally recombination occurs between these linked genes. The higher the frequency of recombination between any two genes, the greater the distance is between them. Recombination frequencies serve as a useful method of mapping genes along a chromosome. No exceptions to Mendellian laws at the level of the gene, but Phenotype ratios are modified

3 RatiosA axa3:1 Aa axa1:1A b a Bxa B9:3:3:1 A ba b a Bxa b1:1:1:1

4 a bc d e f g This map means that there is a 20% recombination frequency between the genes b and c and a 5% recombination frequency between the genes f and g Genes very far apart on the same chromosome will appear to assort independently How many map units between a and f? A-f = 80 cM Recombination freq is not 80% Recombinant/total =

5 Mendel studied 7 traits that assorted independently. The only explanation for this behavior is that the genes controlling these traits are located on different chromosomes. True False Seed color Flower color Pod shape Flower position Stem length Pod color Seed shape

6 The largest distance that can be measured by this technique is 50MU. 50% also indicates NO LINKAGE If two genes are very far apart on the same chromosome, use markers between these genes to more accurately map the genes Therefore when you obtain a recombination frequency of 50% this means that either: __________________ ___________________

7 Interference Interference: this is a phenomenon in which the occurrence of one crossover in a region influences the probability of another crossover occurring in that region. Interference is readily detected genetically. For example, we determined the following map for the genes v ct and cv. Expected double crossovers = product of single crossovers The expected frequency of a double crossover is the product of the two frequencies of single crossovers: DCO= Total progeny = Expected number of DCO is Observed number of DCO = Reduction is because of interference

8 Factors affecting MU In most cases the order of genes revealed by mapping techniques correspond to the order of genes determined by sequencing. In contrast, actual physical distance between genes does not show direct correspondence to map units. -for genes far apart, double, triple etc crossovers affect MU -hotspots of recombination and recombination deserts -Species specific differences Humans 1MU is ~ 1 million bp Yeast 1MU is ~ 2500 bp -Sex specific differences For example markers D12s7 and Pah males recombination frequencies of 9% females recombination frequencies of 22% Extreme example: In Drosophila males, there is no recombination during meiosis

9 Each chromosome consists of one long strand of DNA (about 30 million bp per chromosome arm). The most complete physical map would be a description of sequence of these base pairs (ACTGCCCCGTTTAAATGCGC....) and a description of where each gene resides in this sequence. Cen Tel Recomb Freq Drosophila X Fw Wy FaPa Real distance Fa-Pa = 10 6 bp Fw-Wy = 0.5x10 6 bp

10 Gene order, but not gene distance, is usually consistent between genetic and physical maps. One major reason for this is that the recombination rates are not equal through the length of the chromosome. 2m.u lies between pn and fa- these two genes are located near the telomere 2m.u lie between fw and wy- these two genes are located in the middle of the chromosome What can you conclude about the physical distance between these two sets of genes?

11 Linkage maps in males and females: human chromosome 12.

12 MU to bp Genetic maps are based on recombination frequencies and describe the relative order and relative distance between linked genes. Remember genes reside on chromosomes. So what we would like to know is where are the genes located on the chromosomes 22% Rf = 22MU What does this mean in terms of chromosomes and DNA?

13 Physical maps Physical maps provide information concerning the location of genes on chromosomes Cytological studies have been successfully used to map genes to specific regions of a chromosome. For example in Drosophila in some cells the chromosomes become highly replicated and exhibit very characteristic banding patterns:

14 In situ hybridization Polytene chromosomes Salivary glands Squash on slide Denature/Stain polytene chromosomes label gene probe Hybridize probe to polytene chromosomes Autoradiography

15 Drosophila X chr tip This map is actually very crude. The Drosophila genome consists of about 165 Mega base pairs (165 million bp). This region represents a small fraction (5 to 10 million base pairs).

16 Mitotic recombination Although not as frequent, recombination between homologous chromosomes occurs in mitosis as well as meiosis. This was first discovered in Drosophila by Kurt Stern and now has important implications for the origins of some human cancers Stern made the following cross: ｷ y = yellow body ｷ sn = singed bristles ｷ y+ = normal body ｷ sn+ = normal bristles The F1 females These should be phenotypically normal

17 Mitotic recombination However, some females in a background of normal bristle and normal body color had sectors of singed bristle, normal colored tissue next to sectors of yellow body color normal bristle length. Because these sectors were adjacent to one another, Stern thought that the two spots must be reciprocal products of the same event. That event may be a rare crossover between homologs during the mitotic divisions. y+ sn y sn+

18 A normal mitotic division would occur as follows: Replication Segregation Genotypically identical Daughter cells

19 If you have recombination in mitosis Replication y+ sny sn+ y+ sny sn+ Genotypically different Daughter cells Segregation

20 Mitotic recombination

21 Rb While mitotic recombination is rare, it does occur giving rise to recombinant and wild-type cells. Formation of homozygosity in cells can be carcinogenic if a mutated gene becomes homozygous in somatic cells!!! Retinoblastoma (Rb) is a human tumor that sometimes results from a mitotic recombination event. Rb is a childhood cancer of the eye It occurs from birth to 4 years of age If discovered early enough, treatment is 90% effective

22 Rb There are two forms of Rb: sporadic and hereditary. Hereditary Rb: patients typically develop multiple eye tumors involving both eyes. These tumors develop at an early age. Siblings often develop the same sort of tumors. Sporadic Rb (60% of the cases): The development of the eye tumor is a spontaneous event in the patient with no history of the disease. Tumors develop only in one eye. Occurs later than hereditary Rb

23 Sporadic Retinoblastoma Rb+/Rb+ (phenotypically normal) First point mutation Rb+/Rb- cell (phenotypically normal) Second mutation Rb-/Rb- cell (eye tumor) Hereditary Retinoblastoma Rb+/Rb- (phenotypically normal) One mutation Rb-/Rb- cell (eye tumor)